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dark:focus:bg-gray-800" type="button" ><div class="flex h-full flex-1 items-center justify-center ">Follow</div> <img alt="" class="size-3 flex-none rounded-sm" src="https://cdn-avatars.huggingface.co/v1/production/uploads/651e96991b97c9f33d26bde6/QG_uzb2VV3OeKO0c61ynh.png"> <span class="truncate">Open R1</span></button> <button class="flex h-full items-center border-l pl-1.5 pr-1.5 text-gray-400 hover:bg-gray-50 focus:bg-gray-100 focus:outline-none dark:hover:bg-gray-900 dark:focus:bg-gray-800" title="Show Open R1's followers" type="button">919</button></div> </div> </h1> <div class="mb-3 flex flex-wrap md:mb-4"><div class="mr-1 flex flex-wrap items-center"><span class="mb-1 mr-1 p-1 text-sm leading-tight text-gray-400 md:mb-1.5">Modalities: </span> <a class="mb-1 mr-1 md:mb-1.5 md:mr-1.5 rounded-lg" href="/datasets?modality=modality%3Atext"><div class="tag tag-white "> <svg class="text-red-700 dark:text-red-600" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32"><path fill-rule="evenodd" clip-rule="evenodd" d="M4.619 4.619C2.667 6.573 2.667 9.715 2.667 16s0 9.428 1.952 11.38C6.573 29.333 9.715 29.333 16 29.333s9.428 0 11.38-1.953c1.953-1.95 1.953-5.095 1.953-11.38s0-9.428-1.953-11.381C25.43 2.667 22.285 2.667 16 2.667s-9.428 0-11.381 1.952m8.65 3.714c-.573 0-1.109 0-1.546.066-.495.073-1.003.248-1.41.7-.392.436-.53.956-.59 1.452-.056.464-.056 1.04-.056 1.689V13a1 1 0 1 0 2 0v-.704c0-.724.001-1.176.041-1.505q.015-.15.061-.294a.2.2 0 0 1 .031-.061q0-.003.016-.01a.8.8 0 0 1 .203-.05c.272-.04.654-.043 1.314-.043H15v11.334h-2.333a1 1 0 1 0 0 2H20a1 1 0 0 0 0-2h-3V10.333h1.667c.66 0 1.042.003 1.314.043.123.019.18.04.203.05l.015.009a.2.2 0 0 1 .032.061c.018.05.042.14.061.295.04.329.041.781.041 1.506V13a1 1 0 1 0 2 0v-.76c0-.65 0-1.225-.056-1.69-.06-.495-.198-1.015-.59-1.453-.407-.45-.915-.625-1.41-.698-.437-.067-.973-.067-1.546-.066z" fill="currentColor"></path></svg> <span>Text</span> </div></a> </div><div class="mr-1 flex flex-wrap items-center"><span class="mb-1 mr-1 p-1 text-sm leading-tight text-gray-400 md:mb-1.5">Formats: </span> <a class="mb-1 mr-1 md:mb-1.5 md:mr-1.5 rounded-lg" href="/datasets?format=format%3Aparquet"><div class="tag tag-white "> <svg class="" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 34 34"><path fill-rule="evenodd" clip-rule="evenodd" d="m17.97 18.44-3.98-2.3-7.1 3.78 3.98 2.3 7.1-3.78Zm15.1-1.4-3.99-2.3-16.22 8.63 3.98 2.3 16.22-8.63Zm-5.98-3.45-3.97-2.3-7.1 3.78 3.98 2.3 7.1-3.78Zm-9.94-5.74 3.98 2.3-11.16 5.93L6 13.78l11.16-5.93Zm-13.19 7 3.98 2.3-3.04 1.62-3.98-2.3 3.04-1.61Z" fill="currentColor"></path></svg> <span>parquet</span> </div></a> </div><div class="mr-1 flex flex-wrap items-center"><span class="mb-1 mr-1 p-1 text-sm leading-tight text-gray-400 md:mb-1.5">Languages: </span> <a class="mb-1 mr-1 md:mb-1.5 md:mr-1.5 rounded-lg" href="/datasets?language=language%3Aen"><div class="tag tag-white "> <svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" class="text-green-600/80" preserveAspectRatio="xMidYMid meet" width="1em" height="1em" viewBox="0 0 10 10"><path fill-rule="evenodd" clip-rule="evenodd" d="M0.625 5C0.625 6.16032 1.08594 7.27312 1.90641 8.09359C2.72688 8.91406 3.83968 9.375 5 9.375C6.16032 9.375 7.27312 8.91406 8.09359 8.09359C8.91406 7.27312 9.375 6.16032 9.375 5C9.375 3.83968 8.91406 2.72688 8.09359 1.90641C7.27312 1.08594 6.16032 0.625 5 0.625C3.83968 0.625 2.72688 1.08594 1.90641 1.90641C1.08594 2.72688 0.625 3.83968 0.625 5ZM7.64365 7.48027C7.61734 7.50832 7.59054 7.53598 7.56326 7.56326C7.13828 7.98824 6.61864 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3.6493ZM7.64858 2.52499C7.35446 2.68754 7.0117 2.81868 6.63664 2.91268C6.49676 2.35623 6.29913 1.88209 6.0539 1.53158C6.61864 1.7032 7.13828 2.01176 7.56326 2.43674C7.59224 2.46572 7.62068 2.49514 7.64858 2.52499Z" fill="currentColor"></path></svg> <span>English</span> </div></a> </div><div class="mr-1 flex flex-wrap items-center"><span class="mb-1 mr-1 p-1 text-sm leading-tight text-gray-400 md:mb-1.5">Size: </span> <a class="mb-1 mr-1 md:mb-1.5 md:mr-1.5 rounded-lg" href="/datasets?size_categories=size_categories%3A100K%3Cn%3C1M"><div class="tag tag-white "> <span>100K - 1M</span> </div></a> </div><div class="mr-1 flex flex-wrap items-center"><span class="mb-1 mr-1 p-1 text-sm leading-tight text-gray-400 md:mb-1.5">Libraries: </span> <a class="mb-1 mr-1 md:mb-1.5 md:mr-1.5 rounded-lg" href="/datasets?library=library%3Adatasets"><div class="tag tag-white "><svg class="text-black inline-block text-sm" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" preserveAspectRatio="xMidYMid meet" width="1em" height="1em" viewBox="0 0 95 88"><path fill="#fff" d="M94.25 70.08a8.28 8.28 0 0 1-.43 6.46 10.57 10.57 0 0 1-3 3.6 25.18 25.18 0 0 1-5.7 3.2 65.74 65.74 0 0 1-7.56 2.65 46.67 46.67 0 0 1-11.42 1.68c-5.42.05-10.09-1.23-13.4-4.5a40.4 40.4 0 0 1-10.14.03c-3.34 3.25-7.99 4.52-13.39 4.47a46.82 46.82 0 0 1-11.43-1.68 66.37 66.37 0 0 1-7.55-2.65c-2.28-.98-4.17-2-5.68-3.2a10.5 10.5 0 0 1-3.02-3.6c-.99-2-1.18-4.3-.42-6.46a8.54 8.54 0 0 1-.33-5.63c.25-.95.66-1.83 1.18-2.61a8.67 8.67 0 0 1 2.1-8.47 8.23 8.23 0 0 1 2.82-2.07 41.75 41.75 0 1 1 81.3-.12 8.27 8.27 0 0 1 3.11 2.19 8.7 8.7 0 0 1 2.1 8.47c.52.78.93 1.66 1.18 2.61a8.61 8.61 0 0 1-.32 5.63Z"></path><path fill="#FFD21E" d="M47.21 76.5a34.75 34.75 0 1 0 0-69.5 34.75 34.75 0 0 0 0 69.5Z"></path><path fill="#FF9D0B" d="M81.96 41.75a34.75 34.75 0 1 0-69.5 0 34.75 34.75 0 0 0 69.5 0Zm-73.5 0a38.75 38.75 0 1 1 77.5 0 38.75 38.75 0 0 1-77.5 0Z"></path><path fill="#3A3B45" d="M58.5 32.3c1.28.44 1.78 3.06 3.07 2.38a5 5 0 1 0-6.76-2.07c.61 1.15 2.55-.72 3.7-.32ZM34.95 32.3c-1.28.44-1.79 3.06-3.07 2.38a5 5 0 1 1 6.76-2.07c-.61 1.15-2.56-.72-3.7-.32Z"></path><path fill="#FF323D" d="M46.96 56.29c9.83 0 13-8.76 13-13.26 0-2.34-1.57-1.6-4.09-.36-2.33 1.15-5.46 2.74-8.9 2.74-7.19 0-13-6.88-13-2.38s3.16 13.26 13 13.26Z"></path><path fill="#3A3B45" fill-rule="evenodd" d="M39.43 54a8.7 8.7 0 0 1 5.3-4.49c.4-.12.81.57 1.24 1.28.4.68.82 1.37 1.24 1.37.45 0 .9-.68 1.33-1.35.45-.7.89-1.38 1.32-1.25a8.61 8.61 0 0 1 5 4.17c3.73-2.94 5.1-7.74 5.1-10.7 0-2.34-1.57-1.6-4.09-.36l-.14.07c-2.31 1.15-5.39 2.67-8.77 2.67s-6.45-1.52-8.77-2.67c-2.6-1.29-4.23-2.1-4.23.29 0 3.05 1.46 8.06 5.47 10.97Z" clip-rule="evenodd"></path><path fill="#FF9D0B" d="M70.71 37a3.25 3.25 0 1 0 0-6.5 3.25 3.25 0 0 0 0 6.5ZM24.21 37a3.25 3.25 0 1 0 0-6.5 3.25 3.25 0 0 0 0 6.5ZM17.52 48c-1.62 0-3.06.66-4.07 1.87a5.97 5.97 0 0 0-1.33 3.76 7.1 7.1 0 0 0-1.94-.3c-1.55 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.52-.1.68-.38.86-1.46-8.06-8.22-8.56-11.01-.34-1.9.24-2.85 1.31-2.85Z"></path><path fill="#FFD21E" d="M38.6 76.69c2.75-4.04 2.55-7.07-1.22-10.84-3.78-3.77-5.98-9.3-5.98-9.3s-.82-3.2-2.69-2.9c-1.87.3-3.24 5.08.68 8.01 3.91 2.93-.78 4.92-2.29 2.17-1.5-2.75-5.62-9.82-7.76-11.18-2.13-1.35-3.63-.6-3.13 2.2.5 2.79 9.43 9.55 8.56 11-.87 1.47-3.93-1.71-3.93-1.71s-9.57-8.71-11.66-6.44c-2.08 2.27 1.59 4.17 6.8 7.33 5.23 3.16 5.64 4 4.9 5.2-.75 1.2-12.28-8.53-13.36-4.4-1.08 4.11 11.77 5.3 10.98 8.15-.8 2.85-9.06-5.38-10.74-2.18-1.7 3.21 11.65 6.98 11.76 7.01 4.3 1.12 15.25 3.49 19.08-2.12Z"></path><path fill="#FF9D0B" d="M77.4 48c1.62 0 3.07.66 4.07 1.87a5.97 5.97 0 0 1 1.33 3.76 7.1 7.1 0 0 1 1.95-.3c1.55 0 2.95.59 3.94 1.66a5.8 5.8 0 0 1 .8 7 5.3 5.3 0 0 1 1.78 2.82c.24.9.48 2.8-.8 4.74a5.22 5.22 0 0 1 .37 5.02c-1.02 2.32-3.57 4.14-8.51 6.1-3.08 1.22-5.9 2-5.92 2.01a44.33 44.33 0 0 1-10.93 1.6c-5.86 0-10.05-1.8-12.46-5.34-3.88-5.69-3.33-10.9 1.7-15.92 2.78-2.78 4.63-6.87 5.01-7.77.78-2.66 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Task B-1.3.\n\nA ship traveling along a river has covered $24 \\mathrm{~km}$ upstream and $28 \\mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \\mathrm{~km}$ upstream and $21 \\mathrm{~km}$ downstream, or half an hour more than for traveling $15 \\mathrm{~km}$ upstream and $42 \\mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.\n\nDetermine the speed of the ship in still water and the speed of the river."},"solution":{"kind":"string","value":"## Solution.\n\nLet $t$ be the time required for the boat to travel $24 \\mathrm{~km}$ upstream and $28 \\mathrm{~km}$ downstream, $v_{R}$ the speed of the river, and $v_{B}$ the speed of the boat. When the boat is traveling upstream, its speed is $v_{B}-v_{R}$, and when it is traveling downstream, its speed is $v_{B}+v_{R}$.\n\nSince $t=\\frac{s}{v}$, from the given data, we obtain the following system of equations:\n\n$\\left\\{\\begin{array}{l}t=\\frac{24}{v_{B}-v_{R}}+\\frac{28}{v_{B}+v_{R}} \\\\ t+0.5=\\frac{30}{v_{B}-v_{R}}+\\frac{21}{v_{B}+v_{R}} \\\\ t-0.5=\\frac{15}{v_{B}-v_{R}}+\\frac{42}{v_{B}+v_{R}}\\end{array}\\right.$\n\nBy introducing new variables $x=\\frac{3}{v_{B}-v_{R}}, y=\\frac{7}{v_{B}+v_{R}}$, the system transforms into:\n\n$\\left\\{\\begin{array}{l}t=8 x+4 y \\\\ t+0.5=10 x+3 y \\\\ t-0.5=5 x+6 y\\end{array}\\right.$\n\nSubstituting $t$ from the first equation into the remaining two, we get:\n\n$\\left\\{\\begin{array}{l}8 x+4 y+0.5=10 x+3 y \\\\ 8 x+4 y-0.5=5 x+6 y\\end{array}\\right.$\n\n$\\left\\{\\begin{array}{l}2 x-y=0.5 \\\\ 3 x-2 y=0.5\\end{array}\\right.$\n\nThe solution to the last system is (0.5, 0.5). Then we have:\n\n$\\frac{3}{v_{B}-v_{R}}=0.5$, hence, $v_{B}-v_{R}=6 \\mathrm{~and}$\n\n$\\frac{7}{v_{B}+v_{R}}=0.5$, hence, $v_{B}+v_{R}=14$.\n\nThe speed of the river is $v_{R}=4 \\mathrm{~km} / \\mathrm{h}$, and the speed of the boat is $v_{B}=10 \\mathrm{~km} / \\mathrm{h}$.\n\n## Note:\n\nBy substituting $x=\\frac{1}{v_{B}-v_{R}}, y=\\frac{1}{v_{B}+v_{R}} \\mathrm{~and}$ following the same procedure, the initial system transforms into the system $\\left\\{\\begin{array}{l}6 x-7 y=0.5 \\\\ 9 x-14 y=0.5\\end{array}\\right.$\n\nThe solution to this system is $\\left(\\frac{1}{6}, \\frac{1}{14}\\right)$."},"answer":{"kind":"string","value":"v_{R}=4\\mathrm{~}/\\mathrm{},v_{B}=10\\mathrm{~}/\\mathrm{}"},"problem_type":{"kind":"string","value":"Algebra"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"586fd646-76d6-5070-8c81-9993ab9d8559"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"list like","value":["<think>\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \n\nLet me denote the speed of the ship in still water as \$ v \$ (in km/h) and the speed of the river as \$ r \$ (also in km/h). Then, the upstream speed would be \$ v - r \$, and the downstream speed would be \$ v + r \$.\n\nThe problem mentions three different journeys:\n\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\n\nWait, actually, the problem states: \"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\"\n\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\n\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\n\nTherefore, we can set up two equations based on these time differences.\n\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\n\nTime1 = \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nFor the second journey (30 km up, 21 km down):\n\nTime2 = \$ \\frac{30}{v - r} + \\frac{21}{v + r} \$\n\nAccording to the problem, Time1 = Time2 - 0.5\n\nSimilarly, for the third journey (15 km up, 42 km down):\n\nTime3 = \$ \\frac{15}{v - r} + \\frac{42}{v + r} \$\n\nAnd here, Time1 = Time3 + 0.5\n\nSo, we have two equations:\n\n1. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n2. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nOkay, so now I need to solve these two equations for \$ v \$ and \$ r \$.\n\nLet me first work on the first equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$ from both sides, but maybe it's better to bring all terms to the left side:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{30}{v - r} - \\frac{21}{v + r} + 0.5 = 0 \$\n\nSimplify the terms:\n\nFor the upstream terms: \$ 24 - 30 = -6 \$, so \$ \\frac{-6}{v - r} \$\n\nFor the downstream terms: \$ 28 - 21 = 7 \$, so \$ \\frac{7}{v + r} \$\n\nSo, the equation becomes:\n\n\$ \\frac{-6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{15}{v - r} - \\frac{42}{v + r} - 0.5 = 0 \$\n\nSimplify:\n\nUpstream terms: \$ 24 - 15 = 9 \$, so \$ \\frac{9}{v - r} \$\n\nDownstream terms: \$ 28 - 42 = -14 \$, so \$ \\frac{-14}{v + r} \$\n\nThus:\n\n\$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nNow, we have two equations:\n\n1. \$ -\\frac{6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n2. \$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nLet me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\n1. \$ -6x + 7y + 0.5 = 0 \$ ⇒ \$ -6x + 7y = -0.5 \$\n2. \$ 9x - 14y - 0.5 = 0 \$ ⇒ \$ 9x - 14y = 0.5 \$\n\nNow, we have a system of linear equations:\n\nEquation (1): \$ -6x + 7y = -0.5 \$\n\nEquation (2): \$ 9x - 14y = 0.5 \$\n\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\n\nEquation (1) multiplied by 2: \$ -12x + 14y = -1 \$\n\nNow, add this to Equation (2):\n\n\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \$\n\nSimplify:\n\n\$ -3x = -0.5 \$\n\nThus, \$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \$\n\nBut exact value is \$ x = 1/6 \$\n\nNow, substitute \$ x = 1/6 \$ back into Equation (1):\n\n\$ -6*(1/6) + 7y = -0.5 \$\n\nSimplify:\n\n\$ -1 + 7y = -0.5 \$\n\nAdd 1 to both sides:\n\n7y = 0.5\n\nThus, \$ y = 0.5/7 = 1/14 ≈ 0.0714 \$\n\nSo, \$ x = 1/6 \$ and \$ y = 1/14 \$\n\nRecall that \$ x = \\frac{1}{v - r} = 1/6 \$ ⇒ \$ v - r = 6 \$\n\nSimilarly, \$ y = \\frac{1}{v + r} = 1/14 \$ ⇒ \$ v + r = 14 \$\n\nTherefore, we have two equations:\n\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nAdding these equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nSimplify:\n\n\$ 2v = 20 \$ ⇒ \$ v = 10 \$\n\nSubstituting back into \$ v + r = 14 \$:\n\n\$ 10 + r = 14 \$ ⇒ \$ r = 4 \$\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results by plugging them back into the original time equations.\n\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\n\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\n\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\n\nTherefore, the solution is correct.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$, respectively.\n\nWe are given three journeys with the following time relationships:\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nWe set up the equations for the times taken for each journey:\n1. For the first journey (24 km upstream, 28 km downstream):\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r}\n \\]\n2. For the second journey (30 km upstream, 21 km downstream):\n \\[\n \\frac{30}{v - r} + \\frac{21}{v + r}\n \\]\n3. For the third journey (15 km upstream, 42 km downstream):\n \\[\n \\frac{15}{v - r} + \\frac{42}{v + r}\n \\]\n\nThe time relationships give us the following equations:\n1. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n2. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n\nIntroducing variables \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we transform the equations into:\n1. \\[\n -6x + 7y = -0.5\n \\]\n2. \\[\n 9x - 14y = 0.5\n \\]\n\nSolving this system of linear equations:\n1. Multiply the first equation by 2:\n \\[\n -12x + 14y = -1\n \\]\n2. Add the modified first equation to the second equation:\n \\[\n -3x = -0.5 \\implies x = \\frac{1}{6}\n \\]\n3. Substitute \$ x = \\frac{1}{6} \$ into the first equation:\n \\[\n -6 \\left( \\frac{1}{6} \\right) + 7y = -0.5 \\implies -1 + 7y = -0.5 \\implies 7y = 0.5 \\implies y = \\frac{1}{14}\n \\]\n\nConverting back to \$ v \$ and \$ r \$:\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nSolving these equations:\n1. Adding the two equations:\n \\[\n 2v = 20 \\implies v = 10\n \\]\n2. Substituting \$ v = 10 \$ into \$ v + r = 14 \$:\n \\[\n 10 + r = 14 \\implies r = 4\n \\]\n\nThe speed of the ship in still water is \$\\boxed{10}\$ km/h and the speed of the river is \$\\boxed{4}\$ km/h.","<think>\nOkay, so I need to solve this problem about a ship traveling upstream and downstream. The question is asking for the speed of the ship in still water and the speed of the river. Let me think through this step by step.\n\nFirst, let me parse the problem again. The ship covers different distances upstream and downstream, and the time taken for these journeys is compared. Specifically, it says that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. Additionally, that same journey (24 km up, 28 km down) took half an hour more than traveling 15 km upstream and 42 km downstream. \n\nAll right, so the key here is that the time differences are half an hour in each case. Since the ship and river are moving uniformly, their speeds are constant. Let me denote the speed of the ship in still water as \$ v \$ (km/h) and the speed of the river as \$ r \$ (km/h). Then, when the ship is going upstream, its effective speed is \$ v - r \$, and downstream it's \$ v + r \$.\n\nTime is distance divided by speed, so the time taken for each part of the journey (upstream and downstream) can be calculated by dividing the distance by the respective effective speed. The total time for a journey would be the sum of the time upstream and the time downstream.\n\nLet me write down the equations based on the problem statements. Let's consider the first comparison:\n\n1. The journey of 24 km upstream and 28 km downstream took half an hour less than the journey of 30 km upstream and 21 km downstream.\n\nLet me denote the time for 24 km up and 28 km down as \$ T_1 \$, and the time for 30 km up and 21 km down as \$ T_2 \$. According to the problem, \$ T_1 = T_2 - 0.5 \$ hours.\n\nSimilarly, the second comparison:\n\n2. The same journey (24 km up, 28 km down) took half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nLet me denote the time for 15 km up and 42 km down as \$ T_3 \$. Then, \$ T_1 = T_3 + 0.5 \$ hours.\n\nTherefore, we have two equations:\n\n\\[\n\\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n\\]\n\nand\n\n\\[\n\\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n\\]\n\nHmm, these are the two equations we need to solve. Let me simplify these equations.\n\nStarting with the first equation:\n\nLeft side: \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nRight side: \$ \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract the left side from the right side to set the equation to zero:\n\n\$ \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 - \\left( \\frac{24}{v - r} + \\frac{28}{v + r} \\right) = 0 \$\n\nSimplify term by term:\n\n\$ \\frac{30 - 24}{v - r} + \\frac{21 - 28}{v + r} - 0.5 = 0 \$\n\nWhich simplifies to:\n\n\$ \\frac{6}{v - r} - \\frac{7}{v + r} - 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\nLeft side: \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nRight side: \$ \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nSubtract left side from right side:\n\n\$ \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 - \\left( \\frac{24}{v - r} + \\frac{28}{v + r} \\right) = 0 \$\n\nSimplify term by term:\n\n\$ \\frac{15 - 24}{v - r} + \\frac{42 - 28}{v + r} + 0.5 = 0 \$\n\nWhich simplifies to:\n\n\$ -\\frac{9}{v - r} + \\frac{14}{v + r} + 0.5 = 0 \$\n\nSo now we have two equations:\n\n1. \$ \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5 \$\n2. \$ -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5 \$\n\nLet me write them as:\n\nEquation (1): \$ \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5 \$\n\nEquation (2): \$ -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5 \$\n\nHmm, these are two equations with two variables, \$ \\frac{1}{v - r} \$ and \$ \\frac{1}{v + r} \$. Let me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\nEquation (1): \$ 6x - 7y = 0.5 \$\n\nEquation (2): \$ -9x + 14y = -0.5 \$\n\nNow, I can solve this system of linear equations for x and y. Let me write them again:\n\n1. \$ 6x - 7y = 0.5 \$\n2. \$ -9x + 14y = -0.5 \$\n\nLet me use the elimination method. If I multiply equation (1) by 2, I get:\n\nEquation (1a): \$ 12x - 14y = 1 \$\n\nNow, equation (2) is \$ -9x + 14y = -0.5 \$\n\nIf I add equation (1a) and equation (2), the y terms will cancel out:\n\n\$ (12x - 14y) + (-9x + 14y) = 1 + (-0.5) \$\n\nSimplify:\n\n\$ 3x = 0.5 \$\n\nTherefore, \$ x = 0.5 / 3 = 1/6 \$\n\nNow, substitute x = 1/6 back into equation (1):\n\n\$ 6*(1/6) -7y = 0.5 \$\n\nSimplify:\n\n\$ 1 - 7y = 0.5 \$\n\nSubtract 1:\n\n\$ -7y = -0.5 \$\n\nDivide:\n\n\$ y = (-0.5)/(-7) = 0.5/7 = 1/14 \$\n\nSo, we have x = 1/6 and y = 1/14. Recall that:\n\n\$ x = \\frac{1}{v - r} = 1/6 \$\n\n\$ y = \\frac{1}{v + r} = 1/14 \$\n\nTherefore, we can write:\n\n\$ v - r = 6 \$\n\n\$ v + r = 14 \$\n\nNow, solving these two equations:\n\nAdd the two equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nWhich simplifies to:\n\n\$ 2v = 20 \$\n\nTherefore, \$ v = 10 \$ km/h\n\nSubstituting back into \$ v - r = 6 \$:\n\n\$ 10 - r = 6 \$\n\nThus, \$ r = 10 - 6 = 4 \$ km/h\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results to ensure there are no mistakes.\n\nFirst, check the first journey: 24 km upstream and 28 km downstream.\n\nUpstream speed: 10 - 4 = 6 km/h. Time upstream: 24 / 6 = 4 hours.\n\nDownstream speed: 10 + 4 = 14 km/h. Time downstream: 28 / 14 = 2 hours.\n\nTotal time: 4 + 2 = 6 hours.\n\nSecond journey: 30 km upstream and 21 km downstream.\n\nTime upstream: 30 / 6 = 5 hours.\n\nTime downstream: 21 / 14 = 1.5 hours.\n\nTotal time: 5 + 1.5 = 6.5 hours.\n\nDifference between the two journeys: 6.5 - 6 = 0.5 hours, which is half an hour. That's correct.\n\nThird journey: 15 km upstream and 42 km downstream.\n\nTime upstream: 15 / 6 = 2.5 hours.\n\nTime downstream: 42 / 14 = 3 hours.\n\nTotal time: 2.5 + 3 = 5.5 hours.\n\nDifference between the first journey (6 hours) and this one: 6 - 5.5 = 0.5 hours, which is half an hour more. That also checks out.\n\nTherefore, the solution seems correct. The speed of the ship in still water is 10 km/h and the speed of the river is 4 km/h.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$ respectively.\n\nWe are given two comparisons involving the time taken for different journeys. The first comparison states that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. The second comparison states that the same journey (24 km upstream and 28 km downstream) took half an hour more than traveling 15 km upstream and 42 km downstream.\n\nWe set up the following equations based on the time differences:\n\n1. For the first comparison:\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n Simplifying this, we get:\n \\[\n \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5\n \\]\n\n2. For the second comparison:\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n Simplifying this, we get:\n \\[\n -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5\n \\]\n\nIntroducing \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we convert these equations into a system of linear equations:\n1. \$ 6x - 7y = 0.5 \$\n2. \$ -9x + 14y = -0.5 \$\n\nSolving this system, we multiply the first equation by 2:\n1. \$ 12x - 14y = 1 \$\n\nAdding this to the second equation:\n\\[\n3x = 0.5 \\implies x = \\frac{1}{6}\n\\]\n\nSubstituting \$ x = \\frac{1}{6} \$ back into the first equation:\n\\[\n6 \\left( \\frac{1}{6} \\right) - 7y = 0.5 \\implies 1 - 7y = 0.5 \\implies y = \\frac{1}{14}\n\\]\n\nThus, we have:\n\\[\nv - r = 6 \\quad \\text{and} \\quad v + r = 14\n\\]\n\nSolving these equations:\n\\[\n2v = 20 \\implies v = 10 \\quad \\text{and} \\quad r = 10 - 6 = 4\n\\]\n\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h."],"string":"[\n \"<think>\\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \\n\\nLet me denote the speed of the ship in still water as \\\$ v \\\$ (in km/h) and the speed of the river as \\\$ r \\\$ (also in km/h). Then, the upstream speed would be \\\$ v - r \\\$, and the downstream speed would be \\\$ v + r \\\$.\\n\\nThe problem mentions three different journeys:\\n\\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\\n\\nWait, actually, the problem states: \\\"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\\\"\\n\\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\\n\\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\\n\\nTherefore, we can set up two equations based on these time differences.\\n\\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\\n\\nTime1 = \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$\\n\\nFor the second journey (30 km up, 21 km down):\\n\\nTime2 = \\\$ \\\\frac{30}{v - r} + \\\\frac{21}{v + r} \\\$\\n\\nAccording to the problem, Time1 = Time2 - 0.5\\n\\nSimilarly, for the third journey (15 km up, 42 km down):\\n\\nTime3 = \\\$ \\\\frac{15}{v - r} + \\\\frac{42}{v + r} \\\$\\n\\nAnd here, Time1 = Time3 + 0.5\\n\\nSo, we have two equations:\\n\\n1. \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 \\\$\\n2. \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5 \\\$\\n\\nOkay, so now I need to solve these two equations for \\\$ v \\\$ and \\\$ r \\\$.\\n\\nLet me first work on the first equation:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 \\\$\\n\\nLet me subtract \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$ from both sides, but maybe it's better to bring all terms to the left side:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} - \\\\frac{30}{v - r} - \\\\frac{21}{v + r} + 0.5 = 0 \\\$\\n\\nSimplify the terms:\\n\\nFor the upstream terms: \\\$ 24 - 30 = -6 \\\$, so \\\$ \\\\frac{-6}{v - r} \\\$\\n\\nFor the downstream terms: \\\$ 28 - 21 = 7 \\\$, so \\\$ \\\\frac{7}{v + r} \\\$\\n\\nSo, the equation becomes:\\n\\n\\\$ \\\\frac{-6}{v - r} + \\\\frac{7}{v + r} + 0.5 = 0 \\\$\\n\\nSimilarly, for the second equation:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} - \\\\frac{15}{v - r} - \\\\frac{42}{v + r} - 0.5 = 0 \\\$\\n\\nSimplify:\\n\\nUpstream terms: \\\$ 24 - 15 = 9 \\\$, so \\\$ \\\\frac{9}{v - r} \\\$\\n\\nDownstream terms: \\\$ 28 - 42 = -14 \\\$, so \\\$ \\\\frac{-14}{v + r} \\\$\\n\\nThus:\\n\\n\\\$ \\\\frac{9}{v - r} - \\\\frac{14}{v + r} - 0.5 = 0 \\\$\\n\\nNow, we have two equations:\\n\\n1. \\\$ -\\\\frac{6}{v - r} + \\\\frac{7}{v + r} + 0.5 = 0 \\\$\\n2. \\\$ \\\\frac{9}{v - r} - \\\\frac{14}{v + r} - 0.5 = 0 \\\$\\n\\nLet me denote \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$. Then, the equations become:\\n\\n1. \\\$ -6x + 7y + 0.5 = 0 \\\$ ⇒ \\\$ -6x + 7y = -0.5 \\\$\\n2. \\\$ 9x - 14y - 0.5 = 0 \\\$ ⇒ \\\$ 9x - 14y = 0.5 \\\$\\n\\nNow, we have a system of linear equations:\\n\\nEquation (1): \\\$ -6x + 7y = -0.5 \\\$\\n\\nEquation (2): \\\$ 9x - 14y = 0.5 \\\$\\n\\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\\n\\nEquation (1) multiplied by 2: \\\$ -12x + 14y = -1 \\\$\\n\\nNow, add this to Equation (2):\\n\\n\\\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \\\$\\n\\nSimplify:\\n\\n\\\$ -3x = -0.5 \\\$\\n\\nThus, \\\$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \\\$\\n\\nBut exact value is \\\$ x = 1/6 \\\$\\n\\nNow, substitute \\\$ x = 1/6 \\\$ back into Equation (1):\\n\\n\\\$ -6*(1/6) + 7y = -0.5 \\\$\\n\\nSimplify:\\n\\n\\\$ -1 + 7y = -0.5 \\\$\\n\\nAdd 1 to both sides:\\n\\n7y = 0.5\\n\\nThus, \\\$ y = 0.5/7 = 1/14 ≈ 0.0714 \\\$\\n\\nSo, \\\$ x = 1/6 \\\$ and \\\$ y = 1/14 \\\$\\n\\nRecall that \\\$ x = \\\\frac{1}{v - r} = 1/6 \\\$ ⇒ \\\$ v - r = 6 \\\$\\n\\nSimilarly, \\\$ y = \\\\frac{1}{v + r} = 1/14 \\\$ ⇒ \\\$ v + r = 14 \\\$\\n\\nTherefore, we have two equations:\\n\\n1. \\\$ v - r = 6 \\\$\\n2. \\\$ v + r = 14 \\\$\\n\\nAdding these equations:\\n\\n\\\$ (v - r) + (v + r) = 6 + 14 \\\$\\n\\nSimplify:\\n\\n\\\$ 2v = 20 \\\$ ⇒ \\\$ v = 10 \\\$\\n\\nSubstituting back into \\\$ v + r = 14 \\\$:\\n\\n\\\$ 10 + r = 14 \\\$ ⇒ \\\$ r = 4 \\\$\\n\\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\\n\\nLet me verify these results by plugging them back into the original time equations.\\n\\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\\n\\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\\n\\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\\n\\nTherefore, the solution is correct.\\n\\n**Final Answer**\\nThe speed of the ship in still water is \\\\boxed{10} km/h and the speed of the river is \\\\boxed{4} km/h.\\n</think>\\n\\nLet \\\$ v \\\$ be the speed of the ship in still water (in km/h) and \\\$ r \\\$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \\\$ v - r \\\$ and \\\$ v + r \\\$, respectively.\\n\\nWe are given three journeys with the following time relationships:\\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\\n\\nWe set up the equations for the times taken for each journey:\\n1. For the first journey (24 km upstream, 28 km downstream):\\n \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r}\\n \\\\]\\n2. For the second journey (30 km upstream, 21 km downstream):\\n \\\\[\\n \\\\frac{30}{v - r} + \\\\frac{21}{v + r}\\n \\\\]\\n3. For the third journey (15 km upstream, 42 km downstream):\\n \\\\[\\n \\\\frac{15}{v - r} + \\\\frac{42}{v + r}\\n \\\\]\\n\\nThe time relationships give us the following equations:\\n1. \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5\\n \\\\]\\n2. \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5\\n \\\\]\\n\\nIntroducing variables \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$, we transform the equations into:\\n1. \\\\[\\n -6x + 7y = -0.5\\n \\\\]\\n2. \\\\[\\n 9x - 14y = 0.5\\n \\\\]\\n\\nSolving this system of linear equations:\\n1. Multiply the first equation by 2:\\n \\\\[\\n -12x + 14y = -1\\n \\\\]\\n2. Add the modified first equation to the second equation:\\n \\\\[\\n -3x = -0.5 \\\\implies x = \\\\frac{1}{6}\\n \\\\]\\n3. Substitute \\\$ x = \\\\frac{1}{6} \\\$ into the first equation:\\n \\\\[\\n -6 \\\\left( \\\\frac{1}{6} \\\\right) + 7y = -0.5 \\\\implies -1 + 7y = -0.5 \\\\implies 7y = 0.5 \\\\implies y = \\\\frac{1}{14}\\n \\\\]\\n\\nConverting back to \\\$ v \\\$ and \\\$ r \\\$:\\n1. \\\$ v - r = 6 \\\$\\n2. \\\$ v + r = 14 \\\$\\n\\nSolving these equations:\\n1. Adding the two equations:\\n \\\\[\\n 2v = 20 \\\\implies v = 10\\n \\\\]\\n2. Substituting \\\$ v = 10 \\\$ into \\\$ v + r = 14 \\\$:\\n \\\\[\\n 10 + r = 14 \\\\implies r = 4\\n \\\\]\\n\\nThe speed of the ship in still water is \\\$\\\\boxed{10}\\\$ km/h and the speed of the river is \\\$\\\\boxed{4}\\\$ km/h.\",\n \"<think>\\nOkay, so I need to solve this problem about a ship traveling upstream and downstream. The question is asking for the speed of the ship in still water and the speed of the river. Let me think through this step by step.\\n\\nFirst, let me parse the problem again. The ship covers different distances upstream and downstream, and the time taken for these journeys is compared. Specifically, it says that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. Additionally, that same journey (24 km up, 28 km down) took half an hour more than traveling 15 km upstream and 42 km downstream. \\n\\nAll right, so the key here is that the time differences are half an hour in each case. Since the ship and river are moving uniformly, their speeds are constant. Let me denote the speed of the ship in still water as \\\$ v \\\$ (km/h) and the speed of the river as \\\$ r \\\$ (km/h). Then, when the ship is going upstream, its effective speed is \\\$ v - r \\\$, and downstream it's \\\$ v + r \\\$.\\n\\nTime is distance divided by speed, so the time taken for each part of the journey (upstream and downstream) can be calculated by dividing the distance by the respective effective speed. The total time for a journey would be the sum of the time upstream and the time downstream.\\n\\nLet me write down the equations based on the problem statements. Let's consider the first comparison:\\n\\n1. The journey of 24 km upstream and 28 km downstream took half an hour less than the journey of 30 km upstream and 21 km downstream.\\n\\nLet me denote the time for 24 km up and 28 km down as \\\$ T_1 \\\$, and the time for 30 km up and 21 km down as \\\$ T_2 \\\$. According to the problem, \\\$ T_1 = T_2 - 0.5 \\\$ hours.\\n\\nSimilarly, the second comparison:\\n\\n2. The same journey (24 km up, 28 km down) took half an hour more than the journey of 15 km upstream and 42 km downstream.\\n\\nLet me denote the time for 15 km up and 42 km down as \\\$ T_3 \\\$. Then, \\\$ T_1 = T_3 + 0.5 \\\$ hours.\\n\\nTherefore, we have two equations:\\n\\n\\\\[\\n\\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5\\n\\\\]\\n\\nand\\n\\n\\\\[\\n\\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5\\n\\\\]\\n\\nHmm, these are the two equations we need to solve. Let me simplify these equations.\\n\\nStarting with the first equation:\\n\\nLeft side: \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$\\n\\nRight side: \\\$ \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 \\\$\\n\\nLet me subtract the left side from the right side to set the equation to zero:\\n\\n\\\$ \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 - \\\\left( \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\\right) = 0 \\\$\\n\\nSimplify term by term:\\n\\n\\\$ \\\\frac{30 - 24}{v - r} + \\\\frac{21 - 28}{v + r} - 0.5 = 0 \\\$\\n\\nWhich simplifies to:\\n\\n\\\$ \\\\frac{6}{v - r} - \\\\frac{7}{v + r} - 0.5 = 0 \\\$\\n\\nSimilarly, for the second equation:\\n\\nLeft side: \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$\\n\\nRight side: \\\$ \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5 \\\$\\n\\nSubtract left side from right side:\\n\\n\\\$ \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5 - \\\\left( \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\\right) = 0 \\\$\\n\\nSimplify term by term:\\n\\n\\\$ \\\\frac{15 - 24}{v - r} + \\\\frac{42 - 28}{v + r} + 0.5 = 0 \\\$\\n\\nWhich simplifies to:\\n\\n\\\$ -\\\\frac{9}{v - r} + \\\\frac{14}{v + r} + 0.5 = 0 \\\$\\n\\nSo now we have two equations:\\n\\n1. \\\$ \\\\frac{6}{v - r} - \\\\frac{7}{v + r} = 0.5 \\\$\\n2. \\\$ -\\\\frac{9}{v - r} + \\\\frac{14}{v + r} = -0.5 \\\$\\n\\nLet me write them as:\\n\\nEquation (1): \\\$ \\\\frac{6}{v - r} - \\\\frac{7}{v + r} = 0.5 \\\$\\n\\nEquation (2): \\\$ -\\\\frac{9}{v - r} + \\\\frac{14}{v + r} = -0.5 \\\$\\n\\nHmm, these are two equations with two variables, \\\$ \\\\frac{1}{v - r} \\\$ and \\\$ \\\\frac{1}{v + r} \\\$. Let me denote \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$. Then, the equations become:\\n\\nEquation (1): \\\$ 6x - 7y = 0.5 \\\$\\n\\nEquation (2): \\\$ -9x + 14y = -0.5 \\\$\\n\\nNow, I can solve this system of linear equations for x and y. Let me write them again:\\n\\n1. \\\$ 6x - 7y = 0.5 \\\$\\n2. \\\$ -9x + 14y = -0.5 \\\$\\n\\nLet me use the elimination method. If I multiply equation (1) by 2, I get:\\n\\nEquation (1a): \\\$ 12x - 14y = 1 \\\$\\n\\nNow, equation (2) is \\\$ -9x + 14y = -0.5 \\\$\\n\\nIf I add equation (1a) and equation (2), the y terms will cancel out:\\n\\n\\\$ (12x - 14y) + (-9x + 14y) = 1 + (-0.5) \\\$\\n\\nSimplify:\\n\\n\\\$ 3x = 0.5 \\\$\\n\\nTherefore, \\\$ x = 0.5 / 3 = 1/6 \\\$\\n\\nNow, substitute x = 1/6 back into equation (1):\\n\\n\\\$ 6*(1/6) -7y = 0.5 \\\$\\n\\nSimplify:\\n\\n\\\$ 1 - 7y = 0.5 \\\$\\n\\nSubtract 1:\\n\\n\\\$ -7y = -0.5 \\\$\\n\\nDivide:\\n\\n\\\$ y = (-0.5)/(-7) = 0.5/7 = 1/14 \\\$\\n\\nSo, we have x = 1/6 and y = 1/14. Recall that:\\n\\n\\\$ x = \\\\frac{1}{v - r} = 1/6 \\\$\\n\\n\\\$ y = \\\\frac{1}{v + r} = 1/14 \\\$\\n\\nTherefore, we can write:\\n\\n\\\$ v - r = 6 \\\$\\n\\n\\\$ v + r = 14 \\\$\\n\\nNow, solving these two equations:\\n\\nAdd the two equations:\\n\\n\\\$ (v - r) + (v + r) = 6 + 14 \\\$\\n\\nWhich simplifies to:\\n\\n\\\$ 2v = 20 \\\$\\n\\nTherefore, \\\$ v = 10 \\\$ km/h\\n\\nSubstituting back into \\\$ v - r = 6 \\\$:\\n\\n\\\$ 10 - r = 6 \\\$\\n\\nThus, \\\$ r = 10 - 6 = 4 \\\$ km/h\\n\\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\\n\\nLet me verify these results to ensure there are no mistakes.\\n\\nFirst, check the first journey: 24 km upstream and 28 km downstream.\\n\\nUpstream speed: 10 - 4 = 6 km/h. Time upstream: 24 / 6 = 4 hours.\\n\\nDownstream speed: 10 + 4 = 14 km/h. Time downstream: 28 / 14 = 2 hours.\\n\\nTotal time: 4 + 2 = 6 hours.\\n\\nSecond journey: 30 km upstream and 21 km downstream.\\n\\nTime upstream: 30 / 6 = 5 hours.\\n\\nTime downstream: 21 / 14 = 1.5 hours.\\n\\nTotal time: 5 + 1.5 = 6.5 hours.\\n\\nDifference between the two journeys: 6.5 - 6 = 0.5 hours, which is half an hour. That's correct.\\n\\nThird journey: 15 km upstream and 42 km downstream.\\n\\nTime upstream: 15 / 6 = 2.5 hours.\\n\\nTime downstream: 42 / 14 = 3 hours.\\n\\nTotal time: 2.5 + 3 = 5.5 hours.\\n\\nDifference between the first journey (6 hours) and this one: 6 - 5.5 = 0.5 hours, which is half an hour more. That also checks out.\\n\\nTherefore, the solution seems correct. The speed of the ship in still water is 10 km/h and the speed of the river is 4 km/h.\\n\\n**Final Answer**\\nThe speed of the ship in still water is \\\\boxed{10} km/h and the speed of the river is \\\\boxed{4} km/h.\\n</think>\\n\\nLet \\\$ v \\\$ be the speed of the ship in still water (in km/h) and \\\$ r \\\$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \\\$ v - r \\\$ and \\\$ v + r \\\$ respectively.\\n\\nWe are given two comparisons involving the time taken for different journeys. The first comparison states that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. The second comparison states that the same journey (24 km upstream and 28 km downstream) took half an hour more than traveling 15 km upstream and 42 km downstream.\\n\\nWe set up the following equations based on the time differences:\\n\\n1. For the first comparison:\\n \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5\\n \\\\]\\n Simplifying this, we get:\\n \\\\[\\n \\\\frac{6}{v - r} - \\\\frac{7}{v + r} = 0.5\\n \\\\]\\n\\n2. For the second comparison:\\n \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5\\n \\\\]\\n Simplifying this, we get:\\n \\\\[\\n -\\\\frac{9}{v - r} + \\\\frac{14}{v + r} = -0.5\\n \\\\]\\n\\nIntroducing \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$, we convert these equations into a system of linear equations:\\n1. \\\$ 6x - 7y = 0.5 \\\$\\n2. \\\$ -9x + 14y = -0.5 \\\$\\n\\nSolving this system, we multiply the first equation by 2:\\n1. \\\$ 12x - 14y = 1 \\\$\\n\\nAdding this to the second equation:\\n\\\\[\\n3x = 0.5 \\\\implies x = \\\\frac{1}{6}\\n\\\\]\\n\\nSubstituting \\\$ x = \\\\frac{1}{6} \\\$ back into the first equation:\\n\\\\[\\n6 \\\\left( \\\\frac{1}{6} \\\\right) - 7y = 0.5 \\\\implies 1 - 7y = 0.5 \\\\implies y = \\\\frac{1}{14}\\n\\\\]\\n\\nThus, we have:\\n\\\\[\\nv - r = 6 \\\\quad \\\\text{and} \\\\quad v + r = 14\\n\\\\]\\n\\nSolving these equations:\\n\\\\[\\n2v = 20 \\\\implies v = 10 \\\\quad \\\\text{and} \\\\quad r = 10 - 6 = 4\\n\\\\]\\n\\nThe speed of the ship in still water is \\\\boxed{10} km/h and the speed of the river is \\\\boxed{4} km/h.\"\n]"},"correctness_math_verify":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":1,"string":"1"},"messages":{"kind":"list like","value":[{"content":"## Task B-1.3.\n\nA ship traveling along a river has covered $24 \\mathrm{~km}$ upstream and $28 \\mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \\mathrm{~km}$ upstream and $21 \\mathrm{~km}$ downstream, or half an hour more than for traveling $15 \\mathrm{~km}$ upstream and $42 \\mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.\n\nDetermine the speed of the ship in still water and the speed of the river.","role":"user"},{"content":"<think>\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \n\nLet me denote the speed of the ship in still water as \$ v \$ (in km/h) and the speed of the river as \$ r \$ (also in km/h). Then, the upstream speed would be \$ v - r \$, and the downstream speed would be \$ v + r \$.\n\nThe problem mentions three different journeys:\n\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\n\nWait, actually, the problem states: \"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\"\n\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\n\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\n\nTherefore, we can set up two equations based on these time differences.\n\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\n\nTime1 = \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nFor the second journey (30 km up, 21 km down):\n\nTime2 = \$ \\frac{30}{v - r} + \\frac{21}{v + r} \$\n\nAccording to the problem, Time1 = Time2 - 0.5\n\nSimilarly, for the third journey (15 km up, 42 km down):\n\nTime3 = \$ \\frac{15}{v - r} + \\frac{42}{v + r} \$\n\nAnd here, Time1 = Time3 + 0.5\n\nSo, we have two equations:\n\n1. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n2. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nOkay, so now I need to solve these two equations for \$ v \$ and \$ r \$.\n\nLet me first work on the first equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$ from both sides, but maybe it's better to bring all terms to the left side:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{30}{v - r} - \\frac{21}{v + r} + 0.5 = 0 \$\n\nSimplify the terms:\n\nFor the upstream terms: \$ 24 - 30 = -6 \$, so \$ \\frac{-6}{v - r} \$\n\nFor the downstream terms: \$ 28 - 21 = 7 \$, so \$ \\frac{7}{v + r} \$\n\nSo, the equation becomes:\n\n\$ \\frac{-6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{15}{v - r} - \\frac{42}{v + r} - 0.5 = 0 \$\n\nSimplify:\n\nUpstream terms: \$ 24 - 15 = 9 \$, so \$ \\frac{9}{v - r} \$\n\nDownstream terms: \$ 28 - 42 = -14 \$, so \$ \\frac{-14}{v + r} \$\n\nThus:\n\n\$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nNow, we have two equations:\n\n1. \$ -\\frac{6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n2. \$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nLet me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\n1. \$ -6x + 7y + 0.5 = 0 \$ ⇒ \$ -6x + 7y = -0.5 \$\n2. \$ 9x - 14y - 0.5 = 0 \$ ⇒ \$ 9x - 14y = 0.5 \$\n\nNow, we have a system of linear equations:\n\nEquation (1): \$ -6x + 7y = -0.5 \$\n\nEquation (2): \$ 9x - 14y = 0.5 \$\n\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\n\nEquation (1) multiplied by 2: \$ -12x + 14y = -1 \$\n\nNow, add this to Equation (2):\n\n\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \$\n\nSimplify:\n\n\$ -3x = -0.5 \$\n\nThus, \$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \$\n\nBut exact value is \$ x = 1/6 \$\n\nNow, substitute \$ x = 1/6 \$ back into Equation (1):\n\n\$ -6*(1/6) + 7y = -0.5 \$\n\nSimplify:\n\n\$ -1 + 7y = -0.5 \$\n\nAdd 1 to both sides:\n\n7y = 0.5\n\nThus, \$ y = 0.5/7 = 1/14 ≈ 0.0714 \$\n\nSo, \$ x = 1/6 \$ and \$ y = 1/14 \$\n\nRecall that \$ x = \\frac{1}{v - r} = 1/6 \$ ⇒ \$ v - r = 6 \$\n\nSimilarly, \$ y = \\frac{1}{v + r} = 1/14 \$ ⇒ \$ v + r = 14 \$\n\nTherefore, we have two equations:\n\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nAdding these equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nSimplify:\n\n\$ 2v = 20 \$ ⇒ \$ v = 10 \$\n\nSubstituting back into \$ v + r = 14 \$:\n\n\$ 10 + r = 14 \$ ⇒ \$ r = 4 \$\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results by plugging them back into the original time equations.\n\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\n\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\n\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\n\nTherefore, the solution is correct.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$, respectively.\n\nWe are given three journeys with the following time relationships:\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nWe set up the equations for the times taken for each journey:\n1. For the first journey (24 km upstream, 28 km downstream):\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r}\n \\]\n2. For the second journey (30 km upstream, 21 km downstream):\n \\[\n \\frac{30}{v - r} + \\frac{21}{v + r}\n \\]\n3. For the third journey (15 km upstream, 42 km downstream):\n \\[\n \\frac{15}{v - r} + \\frac{42}{v + r}\n \\]\n\nThe time relationships give us the following equations:\n1. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n2. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n\nIntroducing variables \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we transform the equations into:\n1. \\[\n -6x + 7y = -0.5\n \\]\n2. \\[\n 9x - 14y = 0.5\n \\]\n\nSolving this system of linear equations:\n1. Multiply the first equation by 2:\n \\[\n -12x + 14y = -1\n \\]\n2. Add the modified first equation to the second equation:\n \\[\n -3x = -0.5 \\implies x = \\frac{1}{6}\n \\]\n3. Substitute \$ x = \\frac{1}{6} \$ into the first equation:\n \\[\n -6 \\left( \\frac{1}{6} \\right) + 7y = -0.5 \\implies -1 + 7y = -0.5 \\implies 7y = 0.5 \\implies y = \\frac{1}{14}\n \\]\n\nConverting back to \$ v \$ and \$ r \$:\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nSolving these equations:\n1. Adding the two equations:\n \\[\n 2v = 20 \\implies v = 10\n \\]\n2. Substituting \$ v = 10 \$ into \$ v + r = 14 \$:\n \\[\n 10 + r = 14 \\implies r = 4\n \\]\n\nThe speed of the ship in still water is \$\\boxed{10}\$ km/h and the speed of the river is \$\\boxed{4}\$ km/h.","role":"assistant"}],"string":"[\n {\n \"content\": \"## Task B-1.3.\\n\\nA ship traveling along a river has covered $24 \\\\mathrm{~km}$ upstream and $28 \\\\mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \\\\mathrm{~km}$ upstream and $21 \\\\mathrm{~km}$ downstream, or half an hour more than for traveling $15 \\\\mathrm{~km}$ upstream and $42 \\\\mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.\\n\\nDetermine the speed of the ship in still water and the speed of the river.\",\n \"role\": \"user\"\n },\n {\n \"content\": \"<think>\\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \\n\\nLet me denote the speed of the ship in still water as \\\$ v \\\$ (in km/h) and the speed of the river as \\\$ r \\\$ (also in km/h). Then, the upstream speed would be \\\$ v - r \\\$, and the downstream speed would be \\\$ v + r \\\$.\\n\\nThe problem mentions three different journeys:\\n\\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\\n\\nWait, actually, the problem states: \\\"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\\\"\\n\\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\\n\\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\\n\\nTherefore, we can set up two equations based on these time differences.\\n\\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\\n\\nTime1 = \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$\\n\\nFor the second journey (30 km up, 21 km down):\\n\\nTime2 = \\\$ \\\\frac{30}{v - r} + \\\\frac{21}{v + r} \\\$\\n\\nAccording to the problem, Time1 = Time2 - 0.5\\n\\nSimilarly, for the third journey (15 km up, 42 km down):\\n\\nTime3 = \\\$ \\\\frac{15}{v - r} + \\\\frac{42}{v + r} \\\$\\n\\nAnd here, Time1 = Time3 + 0.5\\n\\nSo, we have two equations:\\n\\n1. \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 \\\$\\n2. \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5 \\\$\\n\\nOkay, so now I need to solve these two equations for \\\$ v \\\$ and \\\$ r \\\$.\\n\\nLet me first work on the first equation:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5 \\\$\\n\\nLet me subtract \\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} \\\$ from both sides, but maybe it's better to bring all terms to the left side:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} - \\\\frac{30}{v - r} - \\\\frac{21}{v + r} + 0.5 = 0 \\\$\\n\\nSimplify the terms:\\n\\nFor the upstream terms: \\\$ 24 - 30 = -6 \\\$, so \\\$ \\\\frac{-6}{v - r} \\\$\\n\\nFor the downstream terms: \\\$ 28 - 21 = 7 \\\$, so \\\$ \\\\frac{7}{v + r} \\\$\\n\\nSo, the equation becomes:\\n\\n\\\$ \\\\frac{-6}{v - r} + \\\\frac{7}{v + r} + 0.5 = 0 \\\$\\n\\nSimilarly, for the second equation:\\n\\n\\\$ \\\\frac{24}{v - r} + \\\\frac{28}{v + r} - \\\\frac{15}{v - r} - \\\\frac{42}{v + r} - 0.5 = 0 \\\$\\n\\nSimplify:\\n\\nUpstream terms: \\\$ 24 - 15 = 9 \\\$, so \\\$ \\\\frac{9}{v - r} \\\$\\n\\nDownstream terms: \\\$ 28 - 42 = -14 \\\$, so \\\$ \\\\frac{-14}{v + r} \\\$\\n\\nThus:\\n\\n\\\$ \\\\frac{9}{v - r} - \\\\frac{14}{v + r} - 0.5 = 0 \\\$\\n\\nNow, we have two equations:\\n\\n1. \\\$ -\\\\frac{6}{v - r} + \\\\frac{7}{v + r} + 0.5 = 0 \\\$\\n2. \\\$ \\\\frac{9}{v - r} - \\\\frac{14}{v + r} - 0.5 = 0 \\\$\\n\\nLet me denote \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$. Then, the equations become:\\n\\n1. \\\$ -6x + 7y + 0.5 = 0 \\\$ ⇒ \\\$ -6x + 7y = -0.5 \\\$\\n2. \\\$ 9x - 14y - 0.5 = 0 \\\$ ⇒ \\\$ 9x - 14y = 0.5 \\\$\\n\\nNow, we have a system of linear equations:\\n\\nEquation (1): \\\$ -6x + 7y = -0.5 \\\$\\n\\nEquation (2): \\\$ 9x - 14y = 0.5 \\\$\\n\\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\\n\\nEquation (1) multiplied by 2: \\\$ -12x + 14y = -1 \\\$\\n\\nNow, add this to Equation (2):\\n\\n\\\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \\\$\\n\\nSimplify:\\n\\n\\\$ -3x = -0.5 \\\$\\n\\nThus, \\\$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \\\$\\n\\nBut exact value is \\\$ x = 1/6 \\\$\\n\\nNow, substitute \\\$ x = 1/6 \\\$ back into Equation (1):\\n\\n\\\$ -6*(1/6) + 7y = -0.5 \\\$\\n\\nSimplify:\\n\\n\\\$ -1 + 7y = -0.5 \\\$\\n\\nAdd 1 to both sides:\\n\\n7y = 0.5\\n\\nThus, \\\$ y = 0.5/7 = 1/14 ≈ 0.0714 \\\$\\n\\nSo, \\\$ x = 1/6 \\\$ and \\\$ y = 1/14 \\\$\\n\\nRecall that \\\$ x = \\\\frac{1}{v - r} = 1/6 \\\$ ⇒ \\\$ v - r = 6 \\\$\\n\\nSimilarly, \\\$ y = \\\\frac{1}{v + r} = 1/14 \\\$ ⇒ \\\$ v + r = 14 \\\$\\n\\nTherefore, we have two equations:\\n\\n1. \\\$ v - r = 6 \\\$\\n2. \\\$ v + r = 14 \\\$\\n\\nAdding these equations:\\n\\n\\\$ (v - r) + (v + r) = 6 + 14 \\\$\\n\\nSimplify:\\n\\n\\\$ 2v = 20 \\\$ ⇒ \\\$ v = 10 \\\$\\n\\nSubstituting back into \\\$ v + r = 14 \\\$:\\n\\n\\\$ 10 + r = 14 \\\$ ⇒ \\\$ r = 4 \\\$\\n\\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\\n\\nLet me verify these results by plugging them back into the original time equations.\\n\\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\\n\\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\\n\\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\\n\\nTherefore, the solution is correct.\\n\\n**Final Answer**\\nThe speed of the ship in still water is \\\\boxed{10} km/h and the speed of the river is \\\\boxed{4} km/h.\\n</think>\\n\\nLet \\\$ v \\\$ be the speed of the ship in still water (in km/h) and \\\$ r \\\$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \\\$ v - r \\\$ and \\\$ v + r \\\$, respectively.\\n\\nWe are given three journeys with the following time relationships:\\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\\n\\nWe set up the equations for the times taken for each journey:\\n1. For the first journey (24 km upstream, 28 km downstream):\\n \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r}\\n \\\\]\\n2. For the second journey (30 km upstream, 21 km downstream):\\n \\\\[\\n \\\\frac{30}{v - r} + \\\\frac{21}{v + r}\\n \\\\]\\n3. For the third journey (15 km upstream, 42 km downstream):\\n \\\\[\\n \\\\frac{15}{v - r} + \\\\frac{42}{v + r}\\n \\\\]\\n\\nThe time relationships give us the following equations:\\n1. \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{30}{v - r} + \\\\frac{21}{v + r} - 0.5\\n \\\\]\\n2. \\\\[\\n \\\\frac{24}{v - r} + \\\\frac{28}{v + r} = \\\\frac{15}{v - r} + \\\\frac{42}{v + r} + 0.5\\n \\\\]\\n\\nIntroducing variables \\\$ x = \\\\frac{1}{v - r} \\\$ and \\\$ y = \\\\frac{1}{v + r} \\\$, we transform the equations into:\\n1. \\\\[\\n -6x + 7y = -0.5\\n \\\\]\\n2. \\\\[\\n 9x - 14y = 0.5\\n \\\\]\\n\\nSolving this system of linear equations:\\n1. Multiply the first equation by 2:\\n \\\\[\\n -12x + 14y = -1\\n \\\\]\\n2. Add the modified first equation to the second equation:\\n \\\\[\\n -3x = -0.5 \\\\implies x = \\\\frac{1}{6}\\n \\\\]\\n3. Substitute \\\$ x = \\\\frac{1}{6} \\\$ into the first equation:\\n \\\\[\\n -6 \\\\left( \\\\frac{1}{6} \\\\right) + 7y = -0.5 \\\\implies -1 + 7y = -0.5 \\\\implies 7y = 0.5 \\\\implies y = \\\\frac{1}{14}\\n \\\\]\\n\\nConverting back to \\\$ v \\\$ and \\\$ r \\\$:\\n1. \\\$ v - r = 6 \\\$\\n2. \\\$ v + r = 14 \\\$\\n\\nSolving these equations:\\n1. Adding the two equations:\\n \\\\[\\n 2v = 20 \\\\implies v = 10\\n \\\\]\\n2. Substituting \\\$ v = 10 \\\$ into \\\$ v + r = 14 \\\$:\\n \\\\[\\n 10 + r = 14 \\\\implies r = 4\\n \\\\]\\n\\nThe speed of the ship in still water is \\\$\\\\boxed{10}\\\$ km/h and the speed of the river is \\\$\\\\boxed{4}\\\$ km/h.\",\n \"role\": \"assistant\"\n }\n]"}}},{"rowIdx":1,"cells":{"problem":{"kind":"string","value":"3. (6 points) A construction company was building a tunnel. When $\\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \\%$ and reduced the working hours to $80 \\%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\\qquad$ days to complete the tunnel."},"solution":{"kind":"string","value":"3. (6 points) A construction company builds a tunnel. When $\\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \\%$ and reducing the daily working hours to $80 \\%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\\qquad$ 180 days.\n\n【Solution】Solution: $\\left(1-\\frac{1}{3}\\right) \\div[(1+20 \\%) \\times 80 \\%]$\n$$\n\\begin{array}{l}\n=\\frac{2}{3} \\div[120 \\% \\times 80 \\%], \\\\\n=\\frac{2}{3} \\div \\frac{24}{25} \\\\\n=\\frac{25}{36} ; \\\\\n185 \\div\\left(\\frac{1}{3}+\\frac{25}{36}\\right) \\\\\n=185 \\div \\frac{37}{36}, \\\\\n=180 \\text { (days). }\n\\end{array}\n$$\n\nAnswer: If the original speed was maintained, it would take 180 days.\nTherefore, the answer is: 180."},"answer":{"kind":"string","value":"180"},"problem_type":{"kind":"string","value":"Algebra"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"1d57f4b3-c516-5522-91cc-a60dd91f6f66"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"list like","value":["<think>\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\n\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\n\nWait, the problem says \"reduced the working hours to 80% of the original.\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\n\nFirst, let's define some variables. Let's let the original construction speed be \"v\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\n\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\n\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\n\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\n\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\n\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\n\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\n\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\n\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\n\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\n\nBut total time is 185 days, so:\n\nTime for Phase 1 + Time for Phase 2 = 185\n\nWhich is:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nWe can solve for R, then T = 1/R is the original total time.\n\nLet me compute that.\n\nFirst, let's write the equation:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nCombine the terms:\n\n[1/3 + (2/3)/0.96] / R = 185\n\nCalculate the numerator:\n\n1/3 + (2/3)/0.96\n\nFirst compute (2/3)/0.96:\n\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\n\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\n\nSo numerator ≈ 1.0277\n\nTherefore, 1.0277 / R = 185\n\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\n\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\n\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\n\nLet me compute 1/3 + (2/3)/0.96 exactly.\n\nFirst, (2/3)/0.96:\n\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\n\nThus, (2/3)*(25/24) = (50/72) = 25/36.\n\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\n\nTherefore, numerator is 37/36.\n\nSo the equation is (37/36)/R = 185\n\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\n\nTherefore, T = 1/R = 180 days.\n\nSo the original time would be 180 days.\n\nWait, so the answer is 180 days? Let me confirm.\n\nIf original time is 180 days, then original rate is 1/180 per day.\n\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\n\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\n\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\n\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\n\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\n\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\n\nTherefore, if they had not used the new equipment, it would have taken 180 days.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original number of days required to complete the tunnel. The original construction speed is \$ \\frac{1}{T} \$ tunnels per day. \n\nWhen \$ \\frac{1}{3} \$ of the tunnel was completed at the original speed, the time taken for this phase is:\n\\[\n\\frac{1/3}{1/T} = \\frac{T}{3} \\text{ days}\n\\]\n\nFor the remaining \$ \\frac{2}{3} \$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \$ 1.2 \\times \\frac{1}{T} \$ per hour, but due to reduced working hours, the effective daily rate becomes:\n\\[\n1.2 \\times \\frac{1}{T} \\times 0.8 = \\frac{0.96}{T} \\text{ tunnels per day}\n\\]\n\nThe time required to complete the remaining \$ \\frac{2}{3} \$ of the tunnel at the new rate is:\n\\[\n\\frac{2/3}{0.96/T} = \\frac{2T}{3 \\times 0.96} = \\frac{2T}{2.88} = \\frac{25T}{36} \\text{ days}\n\\]\n\nThe total time taken with the new equipment is 185 days:\n\\[\n\\frac{T}{3} + \\frac{25T}{36} = 185\n\\]\n\nCombining the terms:\n\\[\n\\frac{12T}{36} + \\frac{25T}{36} = \\frac{37T}{36} = 185\n\\]\n\nSolving for \$ T \$:\n\\[\nT = 185 \\times \\frac{36}{37} = 180\n\\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel.","<think>\nAlright, let's try to solve this problem step by step. So, we have a construction company building a tunnel. They completed the first 1/3 of the tunnel at the original speed, then started using new equipment which increased their speed by 20% and reduced their working hours to 80% of the original. The total time taken was 185 days. We need to find out how many days it would have taken if they didn't use the new equipment and continued at the original speed.\n\nOkay, let's break this down. Let's denote the original speed as S (maybe in terms of tunnel length per day), and the original working hours per day as H. Then, with the new equipment, their speed becomes S + 20% of S, which is 1.2S. Their working hours become 80% of H, which is 0.8H.\n\nWait, but how does working hours per day affect the total days? Hmm. If they reduce the working hours per day, then even if their speed is higher, the amount they can do each day would be speed multiplied by hours. So maybe we need to consider the daily progress before and after the equipment change.\n\nLet me think. Let's assume that originally, without any new equipment, the total tunnel length is L. Then, the total work is L. The original speed is S, so the time to complete the tunnel would be T = L / S days. But they might have a certain number of hours per day, so maybe S is already accounting for hours per day? Wait, the problem mentions \"reduced the working hours to 80% of the original\". So working hours per day is a factor here.\n\nWait, perhaps it's better to model this in terms of man-hours or something? Let's see. Let me try to structure it.\n\nLet me denote:\n\n- Original speed: S (tunnel length per hour)\n\n- Original working hours per day: H\n\nSo originally, the amount of tunnel they can build in one day is S * H.\n\nAfter the new equipment:\n\n- Speed increases by 20%, so new speed is 1.2S\n\n- Working hours per day decrease to 80% of original, so new working hours per day is 0.8H\n\nTherefore, the new daily progress is 1.2S * 0.8H = 0.96SH\n\nWait, that's interesting. So even though speed increased by 20%, the working hours decreased to 80%, so the total daily progress is 0.96 times the original daily progress. So actually, the daily progress decreased by 4%?\n\nBut that seems counterintuitive. Let me check that math again.\n\nOriginal daily progress: S * H\n\nNew daily progress: (1.2S) * (0.8H) = 1.2 * 0.8 * S * H = 0.96SH. Yes, so 0.96 times original daily progress. So actually, despite increasing speed by 20%, the reduction in working hours led to a 4% decrease in daily progress.\n\nHmm, so their daily rate of construction actually decreased slightly when they changed the equipment? But they still managed to complete the tunnel in 185 days. But if they had continued at the original speed and hours, how long would it have taken?\n\nWait, maybe my approach is missing something. Let's see. The problem says \"increased the construction speed by 20% and reduced the working hours to 80% of the original\". So speed per hour is increased by 20%, but they are working fewer hours each day.\n\nAlternatively, maybe \"construction speed\" is already in terms of tunnel length per day? Then, increasing the speed by 20% would mean 1.2 times the original daily speed, but then they reduced the working hours to 80%. Wait, that might be a different interpretation.\n\nWait, the problem is a bit ambiguous. The construction speed could be per hour or per day. The problem says \"increased the construction speed by 20%\". If the original speed was, say, x tunnel length per hour, then a 20% increase would make it 1.2x per hour. Then, if they reduce working hours to 80% of original, the daily progress becomes 1.2x * 0.8H = 0.96xH, which is again 96% of the original daily progress. So the same as before.\n\nAlternatively, if the original speed was given as tunnel length per day, and then they increased that speed by 20%, but then reduced the working hours. Hmm. Wait, the problem says \"increased the construction speed by 20%\". So \"construction speed\" is probably the rate at which they construct, which could be per hour. Then, when they reduce the working hours, their daily progress is (speed per hour)*(hours per day).\n\nSo in any case, the daily progress becomes 0.96 times the original daily progress. So even though they work faster per hour, they work fewer hours each day, resulting in a slight decrease in daily progress.\n\nBut according to the problem, after they started using the new equipment, they managed to complete the tunnel in a total of 185 days. But how does that compare to the original plan?\n\nWait, let's structure the problem.\n\nFirst, they build 1/3 of the tunnel at the original speed. Then, they build the remaining 2/3 with the new equipment. The total time is 185 days. If they hadn't used the new equipment, they would have continued at the original speed for the entire tunnel, and that hypothetical total time is what we need to find.\n\nSo let's denote:\n\nLet T be the original total time needed to build the tunnel without any new equipment. Then, the original daily progress is L / T, where L is the total length.\n\nWait, maybe that's a better way. Let me assume the total length of the tunnel is 1 unit for simplicity. Then, original total time is T days. So original daily progress is 1/T per day.\n\nBut wait, when they start using the new equipment after 1/3 of the tunnel is built, they change their daily progress. Let's figure out the time taken for the first 1/3 and the time taken for the remaining 2/3.\n\nBut how is the daily progress affected?\n\nIf originally, the entire tunnel takes T days, then the original daily progress is 1/T per day. So the first 1/3 of the tunnel would take (1/3) / (1/T) = T/3 days.\n\nAfter that, they switch to the new equipment. With the new equipment, their daily progress is 0.96 * (1/T) per day, as we calculated before? Wait, but no. Wait, if the original daily progress is 1/T per day (since they complete 1 tunnel in T days), then with the new equipment, their daily progress is 0.96*(1/T). Therefore, the remaining 2/3 of the tunnel would take (2/3) / (0.96/T) = (2/3)*(T/0.96) = (2T)/(3*0.96) days.\n\nTherefore, the total time is T/3 + (2T)/(3*0.96) = 185 days.\n\nSo solving for T:\n\nT/3 + (2T)/(3*0.96) = 185\n\nLet me compute this equation.\n\nFirst, factor out T/3:\n\nT/3 [1 + 2/0.96] = 185\n\nCompute 2/0.96: 2 divided by 0.96 is equal to 2.083333...\n\nSo, 1 + 2.083333... = 3.083333...\n\nTherefore, T/3 * 3.083333... = 185\n\nMultiply both sides by 3:\n\nT * 3.083333... = 555\n\nTherefore, T = 555 / 3.083333...\n\nCompute 3.083333... as a fraction. 3.083333... is 3 + 0.083333..., and 0.083333... is 1/12. So 3 + 1/12 = 37/12.\n\nTherefore, T = 555 / (37/12) = 555 * (12/37) = (555/37)*12\n\nCompute 555 divided by 37:\n\n37*15 = 555, so 555/37 = 15\n\nTherefore, T = 15*12 = 180 days.\n\nWait, so the original total time would have been 180 days, and according to the problem, with the new equipment, it took 185 days. That seems like using the new equipment actually made them slower? Because 185 is more than 180. But the problem states that \"as a result, it took a total of 185 days\". Wait, maybe I messed up the daily progress.\n\nWait, hold on. If the original total time is T days, then when they switch to the new equipment after 1/3 of the tunnel is done, which took T/3 days, and then the remaining 2/3 is done at a slower daily rate (0.96 of original daily rate), which would take longer, so total time becomes longer than T? But that contradicts the problem statement, which probably implies that using the new equipment should have made them faster, but maybe not? Wait, maybe not. Wait, let's check.\n\nWait, according to the calculation, when they use the new equipment, their daily progress is 96% of the original, so they actually progress slower each day, so the remaining 2/3 takes longer. So even though they increased speed by 20%, the reduction in working hours caused their daily progress to decrease. Therefore, the total time increased from 180 days to 185 days. But the question is, if they had NOT used the new equipment, how long would it have taken? The answer is 180 days.\n\nBut let's check again if my reasoning is correct.\n\nLet me verify with numbers. Suppose the original total time is 180 days.\n\nFirst 1/3 of the tunnel: 180 / 3 = 60 days.\n\nRemaining 2/3 at original speed: 180 - 60 = 120 days.\n\nBut with the new equipment, daily progress is 0.96 of original. So time taken for remaining 2/3 is 120 / 0.96 = 125 days.\n\nTotal time: 60 + 125 = 185 days. Which matches the problem's statement.\n\nSo that makes sense. Therefore, the original total time is 180 days. Therefore, the answer is 180.\n\nBut let me think again. The problem says \"increased the construction speed by 20% and reduced the working hours to 80% of the original\". So construction speed (per hour) increased by 20%, but working hours per day decreased to 80%. So the daily progress is 1.2 * 0.8 = 0.96 times original. So 96% of original daily progress. Therefore, the remaining 2/3 would take (2/3)/0.96 of the original time. Original time for 2/3 is (2/3)*T, so new time is (2/3)*T /0.96 = (2T)/(3*0.96). Then total time is T/3 + (2T)/(3*0.96) = 185.\n\nSolving for T gives 180. Therefore, the answer is 180 days. That seems correct.\n\nAlternatively, maybe my initial interpretation of speed is wrong. Let's try another approach.\n\nSuppose the original speed is S (length per day). Then, the total length L = S * T, where T is the original total time.\n\nThey complete 1/3 of the tunnel at speed S, which takes (L/3)/S = T/3 days.\n\nThen, they use new equipment, which increases speed by 20%, so new speed is 1.2S. But working hours are reduced to 80% of original. Wait, if the original speed S is length per day, which is dependent on working hours per day. So if they reduce working hours per day to 80%, but speed per hour increases by 20%, then the new speed in terms of length per day would be 1.2 * original speed per hour * 0.8 * original hours per day.\n\nIf original speed per hour is s, then original speed per day is s * H, where H is original hours per day. Then, new speed per day is (1.2s) * (0.8H) = 0.96sH = 0.96S. Therefore, new speed per day is 0.96S.\n\nTherefore, remaining 2/3 of the tunnel would take (2L/3)/(0.96S) = (2/3)*(L/S)/0.96 = (2/3)*T /0.96 = (2T)/(3*0.96) days.\n\nSo total time is T/3 + (2T)/(3*0.96) = 185.\n\nWhich leads to T = 180. So same result.\n\nTherefore, regardless of whether we model speed as per day or per hour, we get the same result. Therefore, the original total time would have been 180 days.\n\nTherefore, the answer is 180, boxed.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original total time needed to complete the tunnel without using the new equipment. The tunnel is divided into two parts: the first \$ \\frac{1}{3} \$ completed at the original speed, and the remaining \$ \\frac{2}{3} \$ completed with the new equipment.\n\n1. **Time to complete the first \$ \\frac{1}{3} \$ of the tunnel:**\n \\[\n \\frac{T}{3} \\text{ days}\n \\]\n\n2. **Effect of new equipment:**\n - Construction speed increases by 20%, so the new speed is \$ 1.2 \$ times the original speed.\n - Working hours are reduced to 80% of the original, so the new working hours are \$ 0.8 \$ times the original.\n\n The new daily progress is \$ 1.2 \\times 0.8 = 0.96 \$ times the original daily progress. Therefore, the time to complete the remaining \$ \\frac{2}{3} \$ of the tunnel with the new equipment is:\n \\[\n \\frac{\\frac{2}{3}T}{0.96} = \\frac{2T}{3 \\times 0.96}\n \\]\n\n3. **Total time with new equipment:**\n \\[\n \\frac{T}{3} + \\frac{2T}{3 \\times 0.96} = 185\n \\]\n\n4. **Solving for \$ T \$:**\n \\[\n \\frac{T}{3} + \\frac{2T}{2.88} = 185\n \\]\n \\[\n \\frac{T}{3} + \\frac{5T}{7.2} = 185\n \\]\n Converting to a common denominator:\n \\[\n \\frac{2.4T}{7.2} + \\frac{5T}{7.2} = 185\n \\]\n \\[\n \\frac{7.4T}{7.2} = 185\n \\]\n \\[\n T = 185 \\times \\frac{7.2}{7.4}\n \\]\n \\[\n T = 180\n \\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel."],"string":"[\n \"<think>\\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\\n\\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\\n\\nWait, the problem says \\\"reduced the working hours to 80% of the original.\\\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\\n\\nFirst, let's define some variables. Let's let the original construction speed be \\\"v\\\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\\n\\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\\n\\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\\n\\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\\n\\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\\n\\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\\n\\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\\n\\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\\n\\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\\n\\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\\n\\nBut total time is 185 days, so:\\n\\nTime for Phase 1 + Time for Phase 2 = 185\\n\\nWhich is:\\n\\n1/(3R) + (2/3)/(0.96R) = 185\\n\\nWe can solve for R, then T = 1/R is the original total time.\\n\\nLet me compute that.\\n\\nFirst, let's write the equation:\\n\\n1/(3R) + (2/3)/(0.96R) = 185\\n\\nCombine the terms:\\n\\n[1/3 + (2/3)/0.96] / R = 185\\n\\nCalculate the numerator:\\n\\n1/3 + (2/3)/0.96\\n\\nFirst compute (2/3)/0.96:\\n\\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\\n\\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\\n\\nSo numerator ≈ 1.0277\\n\\nTherefore, 1.0277 / R = 185\\n\\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\\n\\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\\n\\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\\n\\nLet me compute 1/3 + (2/3)/0.96 exactly.\\n\\nFirst, (2/3)/0.96:\\n\\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\\n\\nThus, (2/3)*(25/24) = (50/72) = 25/36.\\n\\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\\n\\nTherefore, numerator is 37/36.\\n\\nSo the equation is (37/36)/R = 185\\n\\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\\n\\nTherefore, T = 1/R = 180 days.\\n\\nSo the original time would be 180 days.\\n\\nWait, so the answer is 180 days? Let me confirm.\\n\\nIf original time is 180 days, then original rate is 1/180 per day.\\n\\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\\n\\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\\n\\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\\n\\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\\n\\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\\n\\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\\n\\nTherefore, if they had not used the new equipment, it would have taken 180 days.\\n\\n**Final Answer**\\n\\\\boxed{180}\\n</think>\\n\\nLet \\\$ T \\\$ be the original number of days required to complete the tunnel. The original construction speed is \\\$ \\\\frac{1}{T} \\\$ tunnels per day. \\n\\nWhen \\\$ \\\\frac{1}{3} \\\$ of the tunnel was completed at the original speed, the time taken for this phase is:\\n\\\\[\\n\\\\frac{1/3}{1/T} = \\\\frac{T}{3} \\\\text{ days}\\n\\\\]\\n\\nFor the remaining \\\$ \\\\frac{2}{3} \\\$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \\\$ 1.2 \\\\times \\\\frac{1}{T} \\\$ per hour, but due to reduced working hours, the effective daily rate becomes:\\n\\\\[\\n1.2 \\\\times \\\\frac{1}{T} \\\\times 0.8 = \\\\frac{0.96}{T} \\\\text{ tunnels per day}\\n\\\\]\\n\\nThe time required to complete the remaining \\\$ \\\\frac{2}{3} \\\$ of the tunnel at the new rate is:\\n\\\\[\\n\\\\frac{2/3}{0.96/T} = \\\\frac{2T}{3 \\\\times 0.96} = \\\\frac{2T}{2.88} = \\\\frac{25T}{36} \\\\text{ days}\\n\\\\]\\n\\nThe total time taken with the new equipment is 185 days:\\n\\\\[\\n\\\\frac{T}{3} + \\\\frac{25T}{36} = 185\\n\\\\]\\n\\nCombining the terms:\\n\\\\[\\n\\\\frac{12T}{36} + \\\\frac{25T}{36} = \\\\frac{37T}{36} = 185\\n\\\\]\\n\\nSolving for \\\$ T \\\$:\\n\\\\[\\nT = 185 \\\\times \\\\frac{36}{37} = 180\\n\\\\]\\n\\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \\\$\\\\boxed{180}\\\$ days to complete the tunnel.\",\n \"<think>\\nAlright, let's try to solve this problem step by step. So, we have a construction company building a tunnel. They completed the first 1/3 of the tunnel at the original speed, then started using new equipment which increased their speed by 20% and reduced their working hours to 80% of the original. The total time taken was 185 days. We need to find out how many days it would have taken if they didn't use the new equipment and continued at the original speed.\\n\\nOkay, let's break this down. Let's denote the original speed as S (maybe in terms of tunnel length per day), and the original working hours per day as H. Then, with the new equipment, their speed becomes S + 20% of S, which is 1.2S. Their working hours become 80% of H, which is 0.8H.\\n\\nWait, but how does working hours per day affect the total days? Hmm. If they reduce the working hours per day, then even if their speed is higher, the amount they can do each day would be speed multiplied by hours. So maybe we need to consider the daily progress before and after the equipment change.\\n\\nLet me think. Let's assume that originally, without any new equipment, the total tunnel length is L. Then, the total work is L. The original speed is S, so the time to complete the tunnel would be T = L / S days. But they might have a certain number of hours per day, so maybe S is already accounting for hours per day? Wait, the problem mentions \\\"reduced the working hours to 80% of the original\\\". So working hours per day is a factor here.\\n\\nWait, perhaps it's better to model this in terms of man-hours or something? Let's see. Let me try to structure it.\\n\\nLet me denote:\\n\\n- Original speed: S (tunnel length per hour)\\n\\n- Original working hours per day: H\\n\\nSo originally, the amount of tunnel they can build in one day is S * H.\\n\\nAfter the new equipment:\\n\\n- Speed increases by 20%, so new speed is 1.2S\\n\\n- Working hours per day decrease to 80% of original, so new working hours per day is 0.8H\\n\\nTherefore, the new daily progress is 1.2S * 0.8H = 0.96SH\\n\\nWait, that's interesting. So even though speed increased by 20%, the working hours decreased to 80%, so the total daily progress is 0.96 times the original daily progress. So actually, the daily progress decreased by 4%?\\n\\nBut that seems counterintuitive. Let me check that math again.\\n\\nOriginal daily progress: S * H\\n\\nNew daily progress: (1.2S) * (0.8H) = 1.2 * 0.8 * S * H = 0.96SH. Yes, so 0.96 times original daily progress. So actually, despite increasing speed by 20%, the reduction in working hours led to a 4% decrease in daily progress.\\n\\nHmm, so their daily rate of construction actually decreased slightly when they changed the equipment? But they still managed to complete the tunnel in 185 days. But if they had continued at the original speed and hours, how long would it have taken?\\n\\nWait, maybe my approach is missing something. Let's see. The problem says \\\"increased the construction speed by 20% and reduced the working hours to 80% of the original\\\". So speed per hour is increased by 20%, but they are working fewer hours each day.\\n\\nAlternatively, maybe \\\"construction speed\\\" is already in terms of tunnel length per day? Then, increasing the speed by 20% would mean 1.2 times the original daily speed, but then they reduced the working hours to 80%. Wait, that might be a different interpretation.\\n\\nWait, the problem is a bit ambiguous. The construction speed could be per hour or per day. The problem says \\\"increased the construction speed by 20%\\\". If the original speed was, say, x tunnel length per hour, then a 20% increase would make it 1.2x per hour. Then, if they reduce working hours to 80% of original, the daily progress becomes 1.2x * 0.8H = 0.96xH, which is again 96% of the original daily progress. So the same as before.\\n\\nAlternatively, if the original speed was given as tunnel length per day, and then they increased that speed by 20%, but then reduced the working hours. Hmm. Wait, the problem says \\\"increased the construction speed by 20%\\\". So \\\"construction speed\\\" is probably the rate at which they construct, which could be per hour. Then, when they reduce the working hours, their daily progress is (speed per hour)*(hours per day).\\n\\nSo in any case, the daily progress becomes 0.96 times the original daily progress. So even though they work faster per hour, they work fewer hours each day, resulting in a slight decrease in daily progress.\\n\\nBut according to the problem, after they started using the new equipment, they managed to complete the tunnel in a total of 185 days. But how does that compare to the original plan?\\n\\nWait, let's structure the problem.\\n\\nFirst, they build 1/3 of the tunnel at the original speed. Then, they build the remaining 2/3 with the new equipment. The total time is 185 days. If they hadn't used the new equipment, they would have continued at the original speed for the entire tunnel, and that hypothetical total time is what we need to find.\\n\\nSo let's denote:\\n\\nLet T be the original total time needed to build the tunnel without any new equipment. Then, the original daily progress is L / T, where L is the total length.\\n\\nWait, maybe that's a better way. Let me assume the total length of the tunnel is 1 unit for simplicity. Then, original total time is T days. So original daily progress is 1/T per day.\\n\\nBut wait, when they start using the new equipment after 1/3 of the tunnel is built, they change their daily progress. Let's figure out the time taken for the first 1/3 and the time taken for the remaining 2/3.\\n\\nBut how is the daily progress affected?\\n\\nIf originally, the entire tunnel takes T days, then the original daily progress is 1/T per day. So the first 1/3 of the tunnel would take (1/3) / (1/T) = T/3 days.\\n\\nAfter that, they switch to the new equipment. With the new equipment, their daily progress is 0.96 * (1/T) per day, as we calculated before? Wait, but no. Wait, if the original daily progress is 1/T per day (since they complete 1 tunnel in T days), then with the new equipment, their daily progress is 0.96*(1/T). Therefore, the remaining 2/3 of the tunnel would take (2/3) / (0.96/T) = (2/3)*(T/0.96) = (2T)/(3*0.96) days.\\n\\nTherefore, the total time is T/3 + (2T)/(3*0.96) = 185 days.\\n\\nSo solving for T:\\n\\nT/3 + (2T)/(3*0.96) = 185\\n\\nLet me compute this equation.\\n\\nFirst, factor out T/3:\\n\\nT/3 [1 + 2/0.96] = 185\\n\\nCompute 2/0.96: 2 divided by 0.96 is equal to 2.083333...\\n\\nSo, 1 + 2.083333... = 3.083333...\\n\\nTherefore, T/3 * 3.083333... = 185\\n\\nMultiply both sides by 3:\\n\\nT * 3.083333... = 555\\n\\nTherefore, T = 555 / 3.083333...\\n\\nCompute 3.083333... as a fraction. 3.083333... is 3 + 0.083333..., and 0.083333... is 1/12. So 3 + 1/12 = 37/12.\\n\\nTherefore, T = 555 / (37/12) = 555 * (12/37) = (555/37)*12\\n\\nCompute 555 divided by 37:\\n\\n37*15 = 555, so 555/37 = 15\\n\\nTherefore, T = 15*12 = 180 days.\\n\\nWait, so the original total time would have been 180 days, and according to the problem, with the new equipment, it took 185 days. That seems like using the new equipment actually made them slower? Because 185 is more than 180. But the problem states that \\\"as a result, it took a total of 185 days\\\". Wait, maybe I messed up the daily progress.\\n\\nWait, hold on. If the original total time is T days, then when they switch to the new equipment after 1/3 of the tunnel is done, which took T/3 days, and then the remaining 2/3 is done at a slower daily rate (0.96 of original daily rate), which would take longer, so total time becomes longer than T? But that contradicts the problem statement, which probably implies that using the new equipment should have made them faster, but maybe not? Wait, maybe not. Wait, let's check.\\n\\nWait, according to the calculation, when they use the new equipment, their daily progress is 96% of the original, so they actually progress slower each day, so the remaining 2/3 takes longer. So even though they increased speed by 20%, the reduction in working hours caused their daily progress to decrease. Therefore, the total time increased from 180 days to 185 days. But the question is, if they had NOT used the new equipment, how long would it have taken? The answer is 180 days.\\n\\nBut let's check again if my reasoning is correct.\\n\\nLet me verify with numbers. Suppose the original total time is 180 days.\\n\\nFirst 1/3 of the tunnel: 180 / 3 = 60 days.\\n\\nRemaining 2/3 at original speed: 180 - 60 = 120 days.\\n\\nBut with the new equipment, daily progress is 0.96 of original. So time taken for remaining 2/3 is 120 / 0.96 = 125 days.\\n\\nTotal time: 60 + 125 = 185 days. Which matches the problem's statement.\\n\\nSo that makes sense. Therefore, the original total time is 180 days. Therefore, the answer is 180.\\n\\nBut let me think again. The problem says \\\"increased the construction speed by 20% and reduced the working hours to 80% of the original\\\". So construction speed (per hour) increased by 20%, but working hours per day decreased to 80%. So the daily progress is 1.2 * 0.8 = 0.96 times original. So 96% of original daily progress. Therefore, the remaining 2/3 would take (2/3)/0.96 of the original time. Original time for 2/3 is (2/3)*T, so new time is (2/3)*T /0.96 = (2T)/(3*0.96). Then total time is T/3 + (2T)/(3*0.96) = 185.\\n\\nSolving for T gives 180. Therefore, the answer is 180 days. That seems correct.\\n\\nAlternatively, maybe my initial interpretation of speed is wrong. Let's try another approach.\\n\\nSuppose the original speed is S (length per day). Then, the total length L = S * T, where T is the original total time.\\n\\nThey complete 1/3 of the tunnel at speed S, which takes (L/3)/S = T/3 days.\\n\\nThen, they use new equipment, which increases speed by 20%, so new speed is 1.2S. But working hours are reduced to 80% of original. Wait, if the original speed S is length per day, which is dependent on working hours per day. So if they reduce working hours per day to 80%, but speed per hour increases by 20%, then the new speed in terms of length per day would be 1.2 * original speed per hour * 0.8 * original hours per day.\\n\\nIf original speed per hour is s, then original speed per day is s * H, where H is original hours per day. Then, new speed per day is (1.2s) * (0.8H) = 0.96sH = 0.96S. Therefore, new speed per day is 0.96S.\\n\\nTherefore, remaining 2/3 of the tunnel would take (2L/3)/(0.96S) = (2/3)*(L/S)/0.96 = (2/3)*T /0.96 = (2T)/(3*0.96) days.\\n\\nSo total time is T/3 + (2T)/(3*0.96) = 185.\\n\\nWhich leads to T = 180. So same result.\\n\\nTherefore, regardless of whether we model speed as per day or per hour, we get the same result. Therefore, the original total time would have been 180 days.\\n\\nTherefore, the answer is 180, boxed.\\n\\n**Final Answer**\\n\\\\boxed{180}\\n</think>\\n\\nLet \\\$ T \\\$ be the original total time needed to complete the tunnel without using the new equipment. The tunnel is divided into two parts: the first \\\$ \\\\frac{1}{3} \\\$ completed at the original speed, and the remaining \\\$ \\\\frac{2}{3} \\\$ completed with the new equipment.\\n\\n1. **Time to complete the first \\\$ \\\\frac{1}{3} \\\$ of the tunnel:**\\n \\\\[\\n \\\\frac{T}{3} \\\\text{ days}\\n \\\\]\\n\\n2. **Effect of new equipment:**\\n - Construction speed increases by 20%, so the new speed is \\\$ 1.2 \\\$ times the original speed.\\n - Working hours are reduced to 80% of the original, so the new working hours are \\\$ 0.8 \\\$ times the original.\\n\\n The new daily progress is \\\$ 1.2 \\\\times 0.8 = 0.96 \\\$ times the original daily progress. Therefore, the time to complete the remaining \\\$ \\\\frac{2}{3} \\\$ of the tunnel with the new equipment is:\\n \\\\[\\n \\\\frac{\\\\frac{2}{3}T}{0.96} = \\\\frac{2T}{3 \\\\times 0.96}\\n \\\\]\\n\\n3. **Total time with new equipment:**\\n \\\\[\\n \\\\frac{T}{3} + \\\\frac{2T}{3 \\\\times 0.96} = 185\\n \\\\]\\n\\n4. **Solving for \\\$ T \\\$:**\\n \\\\[\\n \\\\frac{T}{3} + \\\\frac{2T}{2.88} = 185\\n \\\\]\\n \\\\[\\n \\\\frac{T}{3} + \\\\frac{5T}{7.2} = 185\\n \\\\]\\n Converting to a common denominator:\\n \\\\[\\n \\\\frac{2.4T}{7.2} + \\\\frac{5T}{7.2} = 185\\n \\\\]\\n \\\\[\\n \\\\frac{7.4T}{7.2} = 185\\n \\\\]\\n \\\\[\\n T = 185 \\\\times \\\\frac{7.2}{7.4}\\n \\\\]\\n \\\\[\\n T = 180\\n \\\\]\\n\\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \\\$\\\\boxed{180}\\\$ days to complete the tunnel.\"\n]"},"correctness_math_verify":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":2,"string":"2"},"messages":{"kind":"list like","value":[{"content":"3. (6 points) A construction company was building a tunnel. When $\\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \\%$ and reduced the working hours to $80 \\%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\\qquad$ days to complete the tunnel.","role":"user"},{"content":"<think>\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\n\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\n\nWait, the problem says \"reduced the working hours to 80% of the original.\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\n\nFirst, let's define some variables. Let's let the original construction speed be \"v\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\n\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\n\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\n\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\n\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\n\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\n\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\n\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\n\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\n\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\n\nBut total time is 185 days, so:\n\nTime for Phase 1 + Time for Phase 2 = 185\n\nWhich is:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nWe can solve for R, then T = 1/R is the original total time.\n\nLet me compute that.\n\nFirst, let's write the equation:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nCombine the terms:\n\n[1/3 + (2/3)/0.96] / R = 185\n\nCalculate the numerator:\n\n1/3 + (2/3)/0.96\n\nFirst compute (2/3)/0.96:\n\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\n\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\n\nSo numerator ≈ 1.0277\n\nTherefore, 1.0277 / R = 185\n\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\n\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\n\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\n\nLet me compute 1/3 + (2/3)/0.96 exactly.\n\nFirst, (2/3)/0.96:\n\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\n\nThus, (2/3)*(25/24) = (50/72) = 25/36.\n\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\n\nTherefore, numerator is 37/36.\n\nSo the equation is (37/36)/R = 185\n\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\n\nTherefore, T = 1/R = 180 days.\n\nSo the original time would be 180 days.\n\nWait, so the answer is 180 days? Let me confirm.\n\nIf original time is 180 days, then original rate is 1/180 per day.\n\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\n\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\n\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\n\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\n\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\n\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\n\nTherefore, if they had not used the new equipment, it would have taken 180 days.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original number of days required to complete the tunnel. The original construction speed is \$ \\frac{1}{T} \$ tunnels per day. \n\nWhen \$ \\frac{1}{3} \$ of the tunnel was completed at the original speed, the time taken for this phase is:\n\\[\n\\frac{1/3}{1/T} = \\frac{T}{3} \\text{ days}\n\\]\n\nFor the remaining \$ \\frac{2}{3} \$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \$ 1.2 \\times \\frac{1}{T} \$ per hour, but due to reduced working hours, the effective daily rate becomes:\n\\[\n1.2 \\times \\frac{1}{T} \\times 0.8 = \\frac{0.96}{T} \\text{ tunnels per day}\n\\]\n\nThe time required to complete the remaining \$ \\frac{2}{3} \$ of the tunnel at the new rate is:\n\\[\n\\frac{2/3}{0.96/T} = \\frac{2T}{3 \\times 0.96} = \\frac{2T}{2.88} = \\frac{25T}{36} \\text{ days}\n\\]\n\nThe total time taken with the new equipment is 185 days:\n\\[\n\\frac{T}{3} + \\frac{25T}{36} = 185\n\\]\n\nCombining the terms:\n\\[\n\\frac{12T}{36} + \\frac{25T}{36} = \\frac{37T}{36} = 185\n\\]\n\nSolving for \$ T \$:\n\\[\nT = 185 \\times \\frac{36}{37} = 180\n\\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel.","role":"assistant"}],"string":"[\n {\n \"content\": \"3. (6 points) A construction company was building a tunnel. When $\\\\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \\\\%$ and reduced the working hours to $80 \\\\%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\\\\qquad$ days to complete the tunnel.\",\n \"role\": \"user\"\n },\n {\n \"content\": \"<think>\\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\\n\\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\\n\\nWait, the problem says \\\"reduced the working hours to 80% of the original.\\\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\\n\\nFirst, let's define some variables. Let's let the original construction speed be \\\"v\\\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\\n\\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\\n\\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\\n\\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\\n\\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\\n\\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\\n\\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\\n\\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\\n\\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\\n\\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\\n\\nBut total time is 185 days, so:\\n\\nTime for Phase 1 + Time for Phase 2 = 185\\n\\nWhich is:\\n\\n1/(3R) + (2/3)/(0.96R) = 185\\n\\nWe can solve for R, then T = 1/R is the original total time.\\n\\nLet me compute that.\\n\\nFirst, let's write the equation:\\n\\n1/(3R) + (2/3)/(0.96R) = 185\\n\\nCombine the terms:\\n\\n[1/3 + (2/3)/0.96] / R = 185\\n\\nCalculate the numerator:\\n\\n1/3 + (2/3)/0.96\\n\\nFirst compute (2/3)/0.96:\\n\\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\\n\\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\\n\\nSo numerator ≈ 1.0277\\n\\nTherefore, 1.0277 / R = 185\\n\\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\\n\\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\\n\\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\\n\\nLet me compute 1/3 + (2/3)/0.96 exactly.\\n\\nFirst, (2/3)/0.96:\\n\\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\\n\\nThus, (2/3)*(25/24) = (50/72) = 25/36.\\n\\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\\n\\nTherefore, numerator is 37/36.\\n\\nSo the equation is (37/36)/R = 185\\n\\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\\n\\nTherefore, T = 1/R = 180 days.\\n\\nSo the original time would be 180 days.\\n\\nWait, so the answer is 180 days? Let me confirm.\\n\\nIf original time is 180 days, then original rate is 1/180 per day.\\n\\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\\n\\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\\n\\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\\n\\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\\n\\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\\n\\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\\n\\nTherefore, if they had not used the new equipment, it would have taken 180 days.\\n\\n**Final Answer**\\n\\\\boxed{180}\\n</think>\\n\\nLet \\\$ T \\\$ be the original number of days required to complete the tunnel. The original construction speed is \\\$ \\\\frac{1}{T} \\\$ tunnels per day. \\n\\nWhen \\\$ \\\\frac{1}{3} \\\$ of the tunnel was completed at the original speed, the time taken for this phase is:\\n\\\\[\\n\\\\frac{1/3}{1/T} = \\\\frac{T}{3} \\\\text{ days}\\n\\\\]\\n\\nFor the remaining \\\$ \\\\frac{2}{3} \\\$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \\\$ 1.2 \\\\times \\\\frac{1}{T} \\\$ per hour, but due to reduced working hours, the effective daily rate becomes:\\n\\\\[\\n1.2 \\\\times \\\\frac{1}{T} \\\\times 0.8 = \\\\frac{0.96}{T} \\\\text{ tunnels per day}\\n\\\\]\\n\\nThe time required to complete the remaining \\\$ \\\\frac{2}{3} \\\$ of the tunnel at the new rate is:\\n\\\\[\\n\\\\frac{2/3}{0.96/T} = \\\\frac{2T}{3 \\\\times 0.96} = \\\\frac{2T}{2.88} = \\\\frac{25T}{36} \\\\text{ days}\\n\\\\]\\n\\nThe total time taken with the new equipment is 185 days:\\n\\\\[\\n\\\\frac{T}{3} + \\\\frac{25T}{36} = 185\\n\\\\]\\n\\nCombining the terms:\\n\\\\[\\n\\\\frac{12T}{36} + \\\\frac{25T}{36} = \\\\frac{37T}{36} = 185\\n\\\\]\\n\\nSolving for \\\$ T \\\$:\\n\\\\[\\nT = 185 \\\\times \\\\frac{36}{37} = 180\\n\\\\]\\n\\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \\\$\\\\boxed{180}\\\$ days to complete the tunnel.\",\n \"role\": \"assistant\"\n }\n]"}}},{"rowIdx":2,"cells":{"problem":{"kind":"truncated","value":"\"Prove that number $1$ can be represented as a sum of a finite number $n$ of real numbers, less than"},"solution":{"kind":"truncated","value":"\"1. **Restate the problem in a more manageable form:**\\n We need to prove that the number \\\$1\\\$ "},"answer":{"kind":"string","value":" 8 "},"problem_type":{"kind":"string","value":"Number Theory"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"aops_forum"},"uuid":{"kind":"string","value":"1cd82e3d-2e7a-5eca-9571-62f5669824c4"},"is_reasoning_complete":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so the problem is to prove that the number 1 can be represented as a sum of a finit"},"correctness_math_verify":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":1,"string":"1"},"messages":{"kind":"truncated","value":"[{\"content\":\"Prove that number $1$ can be represented as a sum of a finite number $n$ of real number"}}},{"rowIdx":3,"cells":{"problem":{"kind":"truncated","value":"\"4. Given the three sides of an obtuse triangle are 3, $4, x$, then the range of values for $x$ is ("},"solution":{"kind":"string","value":"$D$"},"answer":{"kind":"string","value":"D"},"problem_type":{"kind":"string","value":"Geometry"},"question_type":{"kind":"string","value":"MCQ"},"source":{"kind":"string","value":"cn_contest"},"uuid":{"kind":"string","value":"ea24dff6-9e4d-5c55-9ec0-c4d55424997a"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so I have this problem here where I need to find the range of values for x, given t"},"correctness_math_verify":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":2,"string":"2"},"messages":{"kind":"truncated","value":"[{\"content\":\"4. Given the three sides of an obtuse triangle are 3, $4, x$, then the range of values "}}},{"rowIdx":4,"cells":{"problem":{"kind":"truncated","value":"\"1. Solve the equation: $\\\\frac{8 x+13}{3}=\\\\frac{19-12 x}{2}-\\\\left(16 x-\\\\frac{7-4 x}{6}\\\\right)$."},"solution":{"kind":"truncated","value":"\"1. $\\\\frac{8 x+13}{3}=\\\\frac{19-12 x}{2}-\\\\left(16 x-\\\\frac{7-4 x}{6}\\\\right)$\\n\\n$\\\\frac{8 x+13}{3"},"answer":{"kind":"string","value":"\\frac{1}{4}"},"problem_type":{"kind":"string","value":"Algebra"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"a583d65c-9b4c-56ce-8263-47edbc22baa3"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so I need to solve this equation: (8x + 13)/3 = (19 - 12x)/2 - (16x - (7 - 4x)/6). "},"correctness_math_verify":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":2,"string":"2"},"messages":{"kind":"truncated","value":"[{\"content\":\"1. Solve the equation: $\\\\frac{8 x+13}{3}=\\\\frac{19-12 x}{2}-\\\\left(16 x-\\\\frac{7-4 x}{"}}},{"rowIdx":5,"cells":{"problem":{"kind":"truncated","value":"\"A right-angled triangle has side lengths that are integers. What could be the last digit of the are"},"solution":{"kind":"truncated","value":"\"Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to t"},"answer":{"kind":"string","value":"0"},"problem_type":{"kind":"string","value":"Geometry"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"79e837be-2801-5a6a-bc7a-0acee3cdd5fd"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so I need to figure out the possible last digits of the area of a right-angled tria"},"correctness_math_verify":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":2,"string":"2"},"messages":{"kind":"truncated","value":"[{\"content\":\"A right-angled triangle has side lengths that are integers. What could be the last digi"}}},{"rowIdx":6,"cells":{"problem":{"kind":"truncated","value":"\"Let $a_n\\\\ (n\\\\geq 1)$ be the value for which $\\\\int_x^{2x} e^{-t^n}dt\\\\ (x\\\\geq 0)$ is maximal. Fi"},"solution":{"kind":"truncated","value":"\"1. Define the integral \\\$ I_n(x) = \\\\int_x^{2x} e^{-t^n} \\\\, dt \\\$. We need to find the value of "},"answer":{"kind":"string","value":"-\\ln 2"},"problem_type":{"kind":"string","value":"Calculus"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"aops_forum"},"uuid":{"kind":"string","value":"167ecbbb-4270-534f-afa5-75d01d968ab5"},"is_reasoning_complete":{"kind":"list like","value":[true,true,true,true],"string":"[\n true,\n true,\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so I need to find the limit as n approaches infinity of the natural logarithm of a_"},"correctness_math_verify":{"kind":"list like","value":[true,true,true,true],"string":"[\n true,\n true,\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":4,"string":"4"},"messages":{"kind":"truncated","value":"[{\"content\":\"Let $a_n\\\\ (n\\\\geq 1)$ be the value for which $\\\\int_x^{2x} e^{-t^n}dt\\\\ (x\\\\geq 0)$ is"}}},{"rowIdx":7,"cells":{"problem":{"kind":"truncated","value":"\"11. Given that the internal angles $A, B, C$ of $\\\\triangle ABC$ have opposite sides $a, b, c$ resp"},"solution":{"kind":"truncated","value":"\"(1) $\\\\sqrt{3} \\\\sin B \\\\cos \\\\frac{A+B}{2}=\\\\sin C \\\\sin B \\\\Rightarrow \\\\sqrt{3} \\\\sin \\\\frac{C}{"},"answer":{"kind":"string","value":"\\sinA=\\frac{1}{2}or\\sinA=1"},"problem_type":{"kind":"string","value":"Geometry"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"84f224e8-719b-5b42-a410-27816014bdd3"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, so I have this triangle ABC, with angles A, B, C opposite sides a, b, c respectivel"},"correctness_math_verify":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"list like","value":["stop","stop"],"string":"[\n \"stop\",\n \"stop\"\n]"},"correctness_count":{"kind":"number","value":2,"string":"2"},"messages":{"kind":"truncated","value":"[{\"content\":\"11. Given that the internal angles $A, B, C$ of $\\\\triangle ABC$ have opposite sides $a"}}},{"rowIdx":8,"cells":{"problem":{"kind":"truncated","value":"\"Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digits of the nu"},"solution":{"kind":"truncated","value":"\"## Solution.\\n\\nIf between every two digits of the number 14641 we write 2013 zeros, we get a numbe"},"answer":{"kind":"string","value":"x_{1,2}=\\(10^{2014}+1),\\quadx_{3,4}=\\(10^{2014}+1)\\cdoti"},"problem_type":{"kind":"string","value":"Algebra"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"olympiads"},"uuid":{"kind":"string","value":"f769fc87-fed5-5b73-8403-372cdd409800"},"is_reasoning_complete":{"kind":"list like","value":[true,true],"string":"[\n true,\n true\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, let's see. I need to find all complex solutions to the equation x⁴ = n, where n i"},"correctness_math_verify":{"kind":"list like","value":[false,false],"string":"[\n false,\n false\n]"},"correctness_llama":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":1,"string":"1"},"messages":{"kind":"truncated","value":"[{\"content\":\"Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digi"}}},{"rowIdx":9,"cells":{"problem":{"kind":"truncated","value":"\"6. As shown in Figure 2, let $P$ be a point inside the equilateral $\\\\triangle ABC$ with side lengt"},"solution":{"kind":"truncated","value":"\"$$\\n6.11 \\\\sqrt{3} \\\\text {. }\\n$$\\n\\nFor the equilateral $\\\\triangle ABC$ with a height of $12 \\\\t"},"answer":{"kind":"string","value":"11 \\sqrt{3}"},"problem_type":{"kind":"string","value":"Geometry"},"question_type":{"kind":"string","value":"math-word-problem"},"source":{"kind":"string","value":"cn_contest"},"uuid":{"kind":"string","value":"2a4c0374-d9ea-538e-8e0f-0811359d6cf6"},"is_reasoning_complete":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"generations":{"kind":"truncated","value":"[\"<think>\\nOkay, let me try to solve this geometry problem. So, we have an equilateral triangle ABC "},"correctness_math_verify":{"kind":"list like","value":[true,false],"string":"[\n true,\n false\n]"},"correctness_llama":{"kind":"null"},"finish_reasons":{"kind":"null"},"correctness_count":{"kind":"number","value":1,"string":"1"},"messages":{"kind":"truncated","value":"[{\"content\":\"6. As shown in Figure 2, let $P$ be a point inside the equilateral $\\\\triangle ABC$ wit"}}}],"truncated":true},"paginationData":{"pageIndex":0,"numItemsPerPage":100,"numTotalItems":93733,"offset":0,"length":100}},"jwt":"eyJhbGciOiJFZERTQSJ9.eyJyZWFkIjp0cnVlLCJwZXJtaXNzaW9ucyI6eyJyZXBvLmNvbnRlbnQucmVhZCI6dHJ1ZX0sImlhdCI6MTc0MDA0MzYzMSwic3ViIjoiL2RhdGFzZXRzL29wZW4tcjEvT3BlblIxLU1hdGgtMjIwayIsImV4cCI6MTc0MDA0NzIzMSwiaXNzIjoiaHR0cHM6Ly9odWdnaW5nZmFjZS5jbyJ9.dADLFbOHCFpK7tVj6iLH-85MhV1az2zyFqhpPNAp7I66wnkmTijEUgj0ZYnrXs3QuEZnIZ4XR8EF_-tBhe3yAQ","displayUrls":true},"dataset":"open-r1/OpenR1-Math-220k","isGated":false,"isPrivate":false,"hasParquetFormat":true,"author":{"avatarUrl":"https://cdn-avatars.huggingface.co/v1/production/uploads/651e96991b97c9f33d26bde6/QG_uzb2VV3OeKO0c61ynh.png","fullname":"Open R1","name":"open-r1","type":"org","isHf":false,"isMod":false,"isEnterprise":true,"followerCount":919},"compact":true}"><div class="overflow-hidden px-2.5 mx-auto mb-10 rounded-lg border pt-2 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flex-nowrap border-t "><div class="grid flex-1 grid-cols-1 overflow-hidden text-sm md:grid-cols-2 md:place-content-center"><label class="relative block flex-1 px-3 py-2 hover:bg-gray-50 dark:border-gray-850 dark:hover:bg-gray-950 md:border-r max-md:first:border-b md:border-r" title="default"><span class="text-gray-500">Subset (3)</span> <div class="flex items-center whitespace-nowrap"><span class="truncate">default</span> <span class="mx-2 text-gray-500">·</span> <span class="text-gray-500">93.7k rows</span> <svg class="ml-auto min-w-6 pl-2" width="1em" height="1em" viewBox="0 0 12 7" fill="none" xmlns="http://www.w3.org/2000/svg"><path d="M1 1L6 6L11 1" stroke="currentColor"></path></svg></div> <select class="absolute inset-0 z-10 w-full cursor-pointer border-0 bg-white text-base opacity-0"><optgroup label="Subset (3)"><option value="default" selected>default (93.7k rows)</option><option value="all" >all (225k rows)</option><option value="extended" >extended (131k rows)</option></optgroup></select></label> <label class="relative block flex-1 px-3 py-2 hover:bg-gray-50 dark:border-gray-850 dark:hover:bg-gray-900 md:border-r md:border-r-0" title="train"><div class="text-gray-500">Split (1)</div> <div class="flex items-center overflow-hidden whitespace-nowrap"><span class="truncate">train</span> <span class="mx-2 text-gray-500">·</span> <span class="text-gray-500">93.7k rows</span> <svg class="ml-auto min-w-6 pl-2" width="1em" height="1em" viewBox="0 0 12 7" fill="none" xmlns="http://www.w3.org/2000/svg"><path d="M1 1L6 6L11 1" stroke="currentColor"></path></svg></div> <select class="absolute inset-0 z-10 w-full cursor-pointer border-0 bg-white text-base opacity-0"><optgroup label="Split (1)"><option value="train" selected>train (93.7k rows)</option></optgroup></select></label></div> </div> <div class="relative -mx-2.5 flex items-center border-t bg-gradient-to-r text-smd dark:border-gray-900 dark:bg-gray-950 [&:has(:focus)]:from-gray-50 [&:has(:focus)]:to-transparent [&:has(:focus)]:to-20% dark:[&:has(:focus)]:from-gray-900"><form class="flex-1"><svg class="absolute left-3 top-1/2 transform -translate-y-1/2 pointer-events-none text-gray-400" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32"><path d="M30 28.59L22.45 21A11 11 0 1 0 21 22.45L28.59 30zM5 14a9 9 0 1 1 9 9a9 9 0 0 1-9-9z" fill="currentColor"></path></svg> <input disabled class="h-9 w-full border-none bg-transparent px-1 pl-9 pr-3 outline-none placeholder:text-gray-400 " placeholder="Search this dataset" dir="auto"></form> <div class="flex items-center gap-2 px-2 py-1"><button type="button" class="mr-2 flex items-center gap-1 rounded-md border border-yellow-200 bg-yellow-100 pl-0.5 pr-1 text-[.8rem] leading-normal text-gray-700 hover:bg-yellow-200/70 dark:border-orange-500/25 dark:bg-orange-500/20 dark:text-gray-300 dark:hover:brightness-110"><div class="rounded bg-yellow-300 px-1 font-mono text-[.7rem] font-bold text-black dark:bg-yellow-700 dark:text-gray-200">SQL </div> Console </button> </div></div> <div class="-mx-2.5 flex flex-1 flex-col overflow-hidden border-t min-h-64"> <div class="max-h-96 relative overflow-auto"><table class="w-full table-auto rounded-lg font-mono text-xs text-gray-900"><thead class="sticky left-0 right-0 top-0 z-1 bg-white align-top shadow-sm"><tr class="space-y-54 h-full min-w-fit divide-x border-b text-left"><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">problem <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="0" width="11.2" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="13.2" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="26.4" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="39.599999999999994" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="52.8" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="79.19999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="92.39999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="105.6" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="118.8" y="25" width="11.2" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="12.2" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="25.4" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="38.599999999999994" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="51.8" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="78.19999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="91.39999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="104.6" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="117.8" y="0" width="13.2" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">15</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">7.47k</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">solution <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="0" width="11.2" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="13.2" y="19.663270525457236" width="11.2" height="10.336729474542764" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="26.4" y="24.924948504914497" width="11.2" height="5.075051495085503" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="39.599999999999994" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="52.8" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="79.19999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="92.39999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="105.6" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="118.8" y="25" width="11.2" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="12.2" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="25.4" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="38.599999999999994" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="51.8" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="78.19999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="91.39999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="104.6" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="117.8" y="0" width="13.2" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">0</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">13.5k</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">answer <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="0" width="11.2" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="13.2" y="23.846059982534005" width="11.2" height="6.153940017465995" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="26.4" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="39.599999999999994" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="52.8" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="79.19999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="92.39999999999999" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="105.6" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="118.8" y="25" width="11.2" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="12.2" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="25.4" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="38.599999999999994" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="51.8" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="78.19999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="91.39999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="104.6" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="117.8" y="0" width="13.2" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">261</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">problem_type <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">classes</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><defs><clipPath id="rounded-bar"><rect x="0" y="0" width="130" height="8" rx="4"></rect></clipPath><pattern id="hatching" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-400 dark:stroke-gray-500/80" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern><pattern id="hatching-faded" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-100 dark:stroke-gray-500/20" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern></defs><g height="8" style="transform: translateY(20px)" clip-path="url(#rounded-bar)"><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="1" y="0" width="45.18572967898179" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="48.18572967898179" y="0" width="26.648181536918695" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="76.83391121590049" y="0" width="20.465300374467905" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="99.29921159036839" y="0" width="13.29493348127127" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="114.59414507163966" y="0" width="3.9346228116031705" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="120.52876788324284" y="0" width="3.2106515314777084" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="125.73941941472054" y="0" width="2.187105928541709" height="8" fill-opacity="1"></rect></g></g></g><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-white cursor-pointer" x="0" y="0" width="47.18572967898179" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="47.18572967898179" y="0" width="28.648181536918695" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="75.83391121590049" y="0" width="22.465300374467905" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="98.29921159036839" y="0" width="15.29493348127127" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="113.59414507163966" y="0" width="5.9346228116031705" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="119.52876788324284" y="0" width="5.210651531477708" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="124.73941941472054" y="0" width="4.187105928541709" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="128.92652534326226" y="0" width="1.073474656737755" height="28" fill-opacity="0"></rect></g></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 max-w-full overflow-hidden text-ellipsis whitespace-nowrap">8 values</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">question_type <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">classes</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><defs><clipPath id="rounded-bar"><rect x="0" y="0" width="130" height="8" rx="4"></rect></clipPath><pattern id="hatching" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-400 dark:stroke-gray-500/80" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern><pattern id="hatching-faded" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-100 dark:stroke-gray-500/20" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern></defs><g height="8" style="transform: translateY(20px)" clip-path="url(#rounded-bar)"><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="1" y="0" width="105.64981383290838" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="108.64981383290838" y="0" width="20.13521385211185" height="8" fill-opacity="1"></rect></g></g></g><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-white cursor-pointer" x="0" y="0" width="107.64981383290838" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="107.64981383290838" y="0" width="22.13521385211185" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="129.78502768502022" y="0" width="0.21497231497978297" height="28" fill-opacity="0"></rect></g></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 max-w-full overflow-hidden text-ellipsis whitespace-nowrap">3 values</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">source <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">classes</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><defs><clipPath id="rounded-bar"><rect x="0" y="0" width="130" height="8" rx="4"></rect></clipPath><pattern id="hatching" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-400 dark:stroke-gray-500/80" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern><pattern id="hatching-faded" patternUnits="userSpaceOnUse" patternTransform="rotate(-45)" height="1" width="5"><line y1="0" class="stroke-gray-100 dark:stroke-gray-500/20" stroke-width="3" y2="1" x1="2" x2="2"></line></pattern></defs><g height="8" style="transform: translateY(20px)" clip-path="url(#rounded-bar)"><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="1" y="0" width="92.43387067521577" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="95.43387067521577" y="0" width="14.706816169332036" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="112.14068684454782" y="0" width="9.573832054879285" height="8" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" x="123.7145188994271" y="0" width="2.7196825024271067" height="8" fill-opacity="1"></rect></g></g></g><g style="transform: scaleX(1.0153846153846153) translateX(-1px)"><g><rect class="fill-white cursor-pointer" x="0" y="0" width="94.43387067521577" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="94.43387067521577" y="0" width="16.706816169332036" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="111.14068684454782" y="0" width="11.573832054879285" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="122.7145188994271" y="0" width="4.719682502427107" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="127.43420140185421" y="0" width="1.1178560378948716" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="128.5520574397491" y="0" width="0.9209136590101672" height="28" fill-opacity="0"></rect><rect class="fill-white cursor-pointer" x="129.47297109875925" y="0" width="0.5270289012407583" height="28" fill-opacity="0"></rect></g></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 max-w-full overflow-hidden text-ellipsis whitespace-nowrap">7 values</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">uuid <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>string</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="25" width="11.2" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="13.2" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="26.4" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="39.599999999999994" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="52.8" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="79.19999999999999" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="92.39999999999999" y="26" width="11.2" height="4" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="105.6" y="0" width="11.2" height="30" fill-opacity="1"></rect><rect class=" fill-gray-200 dark:fill-gray-500/80" rx="2" x="118.8" y="25" width="11.2" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="12.2" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="25.4" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="38.599999999999994" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="51.8" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="78.19999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="91.39999999999999" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="104.6" y="0" width="13.2" height="30" fill-opacity="0"></rect><rect class="fill-gray-200 cursor-pointer" x="117.8" y="0" width="13.2" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 53.4px">3</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 13.2px; max-width: 53.4px">36</div> <div class="absolute -translate-x-1/2" style="left: 124.39999999999999px">⌀</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">is_reasoning_complete <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>sequence</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="22" y="0" width="20" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="44" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="24.910690060595098" width="20" height="5.089309939404902" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="88" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="110" y="25" width="20" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="21" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="87" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="109" y="0" width="22" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">6</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">generations <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>sequence</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="22" y="0" width="20" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="44" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="24.910690060595098" width="20" height="5.089309939404902" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="88" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="110" y="25" width="20" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="21" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="87" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="109" y="0" width="22" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">6</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">correctness_math_verify <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>sequence</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="22" y="0" width="20" height="30" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="44" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="24.910690060595098" width="20" height="5.089309939404902" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="88" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="110" y="25" width="20" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="21" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="87" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="109" y="0" width="22" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">6</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">correctness_llama <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>sequence</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="25" width="42" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="44" y="15.145301508309526" width="42" height="14.854698491690474" fill-opacity="1"></rect><rect class=" fill-gray-200 dark:fill-gray-500/80" rx="2" x="88" y="0" width="42" height="30" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="44" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="44" height="30" fill-opacity="0"></rect><rect class="fill-gray-200 cursor-pointer" x="87" y="0" width="44" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 38px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 44px; max-width: 38px">2</div> <div class="absolute -translate-x-1/2" style="left: 109px">⌀</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">finish_reasons <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>sequence</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="21.001707006369426" width="20" height="8.998292993630574" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="22" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="44" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="66" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="88" y="25" width="20" height="5" fill-opacity="1"></rect><rect class=" fill-gray-200 dark:fill-gray-500/80" rx="2" x="110" y="0" width="20" height="30" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="21" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="87" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-gray-200 cursor-pointer" x="109" y="0" width="22" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 49px">2</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 22px; max-width: 49px">6</div> <div class="absolute -translate-x-1/2" style="left: 120px">⌀</div></div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">correctness_count <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>int64</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="0" y="15.647574383024525" width="20" height="14.352425616975475" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="22" y="0" width="20" height="30" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="44" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="66" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="88" y="25" width="20" height="5" fill-opacity="1"></rect><rect class="fill-indigo-500 dark:fill-indigo-600/80" rx="2" x="110" y="25" width="20" height="5" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="21" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="43" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="65" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="87" y="0" width="22" height="30" fill-opacity="0"></rect><rect class="fill-indigo-500 cursor-pointer" x="109" y="0" width="22" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">1</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">6</div> </div></div></div></div></th><th class="h-full max-w-sm p-2 text-left "><div class="flex h-full flex-col flex-nowrap justify-between"><div><div class="flex items-center justify-between">messages <form class="flex flex-col"><button id="asc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="-rotate-180 transform text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button> <button id="desc" class="-mr-1 ml-2 h-[0.4rem] w-[0.8rem] transition ease-in-out"><svg class="text-gray-300 hover:text-gray-500" xmlns="http://www.w3.org/2000/svg" viewBox="0 64 256 128" fill="currentColor" aria-hidden="true"><path d="M213.65674,101.657l-80,79.99976a7.99945,7.99945,0,0,1-11.31348,0l-80-79.99976A8,8,0,0,1,48,88H208a8,8,0,0,1,5.65674,13.657Z"></path></svg></button></form></div> <div class="mb-2 whitespace-nowrap text-xs font-normal text-gray-500"><span>list</span><span class="italic text-gray-400 before:mx-1 before:content-['·']">lengths</span></div></div> <div><div class="" style="height: 40px; padding-top: 2px"><svg width="130" height="28"><g><rect class="fill-gray-400 dark:fill-gray-500/80" rx="2" x="0" y="0" width="130" height="30" fill-opacity="1"></rect></g><rect class="fill-white dark:fill-gray-900" x="0" y="26" width="130" height="2" stroke-opacity="1"></rect><line class="stroke-gray-100 dark:stroke-gray-500/20" x1="0" y1="27.5" x2="130" y2="27.5" stroke-opacity="1"></line><g><rect class="fill-indigo-500 cursor-pointer" x="-1" y="0" width="132" height="30" fill-opacity="0"></rect></g></svg> <div class="relative font-light text-gray-400" style="height: 10px; width: 130px;"><div class="absolute left-0 overflow-hidden text-ellipsis whitespace-nowrap" style="max-width: 60px">2</div> <div class="absolute overflow-hidden text-ellipsis whitespace-nowrap" style="right: 0px; max-width: 60px">2</div> </div></div></div></div></th></tr></thead> <tbody class="h-16 overflow-scroll"><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="0"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">## Task B-1.3. A ship traveling along a river has covered $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream, or half an hour more than for traveling $15 \mathrm{~km}$ upstream and $42 \mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly. Determine the speed of the ship in still water and the speed of the river.</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">## Solution. Let $t$ be the time required for the boat to travel $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream, $v_{R}$ the speed of the river, and $v_{B}$ the speed of the boat. When the boat is traveling upstream, its speed is $v_{B}-v_{R}$, and when it is traveling downstream, its speed is $v_{B}+v_{R}$. Since $t=\frac{s}{v}$, from the given data, we obtain the following system of equations: $\left\{\begin{array}{l}t=\frac{24}{v_{B}-v_{R}}+\frac{28}{v_{B}+v_{R}} \\ t+0.5=\frac{30}{v_{B}-v_{R}}+\frac{21}{v_{B}+v_{R}} \\ t-0.5=\frac{15}{v_{B}-v_{R}}+\frac{42}{v_{B}+v_{R}}\end{array}\right.$ By introducing new variables $x=\frac{3}{v_{B}-v_{R}}, y=\frac{7}{v_{B}+v_{R}}$, the system transforms into: $\left\{\begin{array}{l}t=8 x+4 y \\ t+0.5=10 x+3 y \\ t-0.5=5 x+6 y\end{array}\right.$ Substituting $t$ from the first equation into the remaining two, we get: $\left\{\begin{array}{l}8 x+4 y+0.5=10 x+3 y \\ 8 x+4 y-0.5=5 x+6 y\end{array}\right.$ $\left\{\begin{array}{l}2 x-y=0.5 \\ 3 x-2 y=0.5\end{array}\right.$ The solution to the last system is (0.5, 0.5). Then we have: $\frac{3}{v_{B}-v_{R}}=0.5$, hence, $v_{B}-v_{R}=6 \mathrm{~and}$ $\frac{7}{v_{B}+v_{R}}=0.5$, hence, $v_{B}+v_{R}=14$. The speed of the river is $v_{R}=4 \mathrm{~km} / \mathrm{h}$, and the speed of the boat is $v_{B}=10 \mathrm{~km} / \mathrm{h}$. ## Note: By substituting $x=\frac{1}{v_{B}-v_{R}}, y=\frac{1}{v_{B}+v_{R}} \mathrm{~and}$ following the same procedure, the initial system transforms into the system $\left\{\begin{array}{l}6 x-7 y=0.5 \\ 9 x-14 y=0.5\end{array}\right.$ The solution to this system is $\left(\frac{1}{6}, \frac{1}{14}\right)$.</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">v_{R}=4\mathrm{~}/\mathrm{},v_{B}=10\mathrm{~}/\mathrm{}</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Algebra</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">586fd646-76d6-5070-8c81-9993ab9d8559</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ "<think>\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \n\nLet me denote the speed of the ship in still water as \$ v \$ (in km/h) and the speed of the river as \$ r \$ (also in km/h). Then, the upstream speed would be \$ v - r \$, and the downstream speed would be \$ v + r \$.\n\nThe problem mentions three different journeys:\n\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\n\nWait, actually, the problem states: \"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\"\n\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\n\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\n\nTherefore, we can set up two equations based on these time differences.\n\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\n\nTime1 = \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nFor the second journey (30 km up, 21 km down):\n\nTime2 = \$ \\frac{30}{v - r} + \\frac{21}{v + r} \$\n\nAccording to the problem, Time1 = Time2 - 0.5\n\nSimilarly, for the third journey (15 km up, 42 km down):\n\nTime3 = \$ \\frac{15}{v - r} + \\frac{42}{v + r} \$\n\nAnd here, Time1 = Time3 + 0.5\n\nSo, we have two equations:\n\n1. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n2. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nOkay, so now I need to solve these two equations for \$ v \$ and \$ r \$.\n\nLet me first work on the first equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$ from both sides, but maybe it's better to bring all terms to the left side:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{30}{v - r} - \\frac{21}{v + r} + 0.5 = 0 \$\n\nSimplify the terms:\n\nFor the upstream terms: \$ 24 - 30 = -6 \$, so \$ \\frac{-6}{v - r} \$\n\nFor the downstream terms: \$ 28 - 21 = 7 \$, so \$ \\frac{7}{v + r} \$\n\nSo, the equation becomes:\n\n\$ \\frac{-6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{15}{v - r} - \\frac{42}{v + r} - 0.5 = 0 \$\n\nSimplify:\n\nUpstream terms: \$ 24 - 15 = 9 \$, so \$ \\frac{9}{v - r} \$\n\nDownstream terms: \$ 28 - 42 = -14 \$, so \$ \\frac{-14}{v + r} \$\n\nThus:\n\n\$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nNow, we have two equations:\n\n1. \$ -\\frac{6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n2. \$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nLet me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\n1. \$ -6x + 7y + 0.5 = 0 \$ ⇒ \$ -6x + 7y = -0.5 \$\n2. \$ 9x - 14y - 0.5 = 0 \$ ⇒ \$ 9x - 14y = 0.5 \$\n\nNow, we have a system of linear equations:\n\nEquation (1): \$ -6x + 7y = -0.5 \$\n\nEquation (2): \$ 9x - 14y = 0.5 \$\n\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\n\nEquation (1) multiplied by 2: \$ -12x + 14y = -1 \$\n\nNow, add this to Equation (2):\n\n\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \$\n\nSimplify:\n\n\$ -3x = -0.5 \$\n\nThus, \$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \$\n\nBut exact value is \$ x = 1/6 \$\n\nNow, substitute \$ x = 1/6 \$ back into Equation (1):\n\n\$ -6*(1/6) + 7y = -0.5 \$\n\nSimplify:\n\n\$ -1 + 7y = -0.5 \$\n\nAdd 1 to both sides:\n\n7y = 0.5\n\nThus, \$ y = 0.5/7 = 1/14 ≈ 0.0714 \$\n\nSo, \$ x = 1/6 \$ and \$ y = 1/14 \$\n\nRecall that \$ x = \\frac{1}{v - r} = 1/6 \$ ⇒ \$ v - r = 6 \$\n\nSimilarly, \$ y = \\frac{1}{v + r} = 1/14 \$ ⇒ \$ v + r = 14 \$\n\nTherefore, we have two equations:\n\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nAdding these equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nSimplify:\n\n\$ 2v = 20 \$ ⇒ \$ v = 10 \$\n\nSubstituting back into \$ v + r = 14 \$:\n\n\$ 10 + r = 14 \$ ⇒ \$ r = 4 \$\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results by plugging them back into the original time equations.\n\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\n\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\n\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\n\nTherefore, the solution is correct.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$, respectively.\n\nWe are given three journeys with the following time relationships:\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nWe set up the equations for the times taken for each journey:\n1. For the first journey (24 km upstream, 28 km downstream):\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r}\n \\]\n2. For the second journey (30 km upstream, 21 km downstream):\n \\[\n \\frac{30}{v - r} + \\frac{21}{v + r}\n \\]\n3. For the third journey (15 km upstream, 42 km downstream):\n \\[\n \\frac{15}{v - r} + \\frac{42}{v + r}\n \\]\n\nThe time relationships give us the following equations:\n1. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n2. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n\nIntroducing variables \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we transform the equations into:\n1. \\[\n -6x + 7y = -0.5\n \\]\n2. \\[\n 9x - 14y = 0.5\n \\]\n\nSolving this system of linear equations:\n1. Multiply the first equation by 2:\n \\[\n -12x + 14y = -1\n \\]\n2. Add the modified first equation to the second equation:\n \\[\n -3x = -0.5 \\implies x = \\frac{1}{6}\n \\]\n3. Substitute \$ x = \\frac{1}{6} \$ into the first equation:\n \\[\n -6 \\left( \\frac{1}{6} \\right) + 7y = -0.5 \\implies -1 + 7y = -0.5 \\implies 7y = 0.5 \\implies y = \\frac{1}{14}\n \\]\n\nConverting back to \$ v \$ and \$ r \$:\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nSolving these equations:\n1. Adding the two equations:\n \\[\n 2v = 20 \\implies v = 10\n \\]\n2. Substituting \$ v = 10 \$ into \$ v + r = 14 \$:\n \\[\n 10 + r = 14 \\implies r = 4\n \\]\n\nThe speed of the ship in still water is \$\\boxed{10}\$ km/h and the speed of the river is \$\\boxed{4}\$ km/h.", "<think>\nOkay, so I need to solve this problem about a ship traveling upstream and downstream. The question is asking for the speed of the ship in still water and the speed of the river. Let me think through this step by step.\n\nFirst, let me parse the problem again. The ship covers different distances upstream and downstream, and the time taken for these journeys is compared. Specifically, it says that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. Additionally, that same journey (24 km up, 28 km down) took half an hour more than traveling 15 km upstream and 42 km downstream. \n\nAll right, so the key here is that the time differences are half an hour in each case. Since the ship and river are moving uniformly, their speeds are constant. Let me denote the speed of the ship in still water as \$ v \$ (km/h) and the speed of the river as \$ r \$ (km/h). Then, when the ship is going upstream, its effective speed is \$ v - r \$, and downstream it's \$ v + r \$.\n\nTime is distance divided by speed, so the time taken for each part of the journey (upstream and downstream) can be calculated by dividing the distance by the respective effective speed. The total time for a journey would be the sum of the time upstream and the time downstream.\n\nLet me write down the equations based on the problem statements. Let's consider the first comparison:\n\n1. The journey of 24 km upstream and 28 km downstream took half an hour less than the journey of 30 km upstream and 21 km downstream.\n\nLet me denote the time for 24 km up and 28 km down as \$ T_1 \$, and the time for 30 km up and 21 km down as \$ T_2 \$. According to the problem, \$ T_1 = T_2 - 0.5 \$ hours.\n\nSimilarly, the second comparison:\n\n2. The same journey (24 km up, 28 km down) took half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nLet me denote the time for 15 km up and 42 km down as \$ T_3 \$. Then, \$ T_1 = T_3 + 0.5 \$ hours.\n\nTherefore, we have two equations:\n\n\\[\n\\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n\\]\n\nand\n\n\\[\n\\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n\\]\n\nHmm, these are the two equations we need to solve. Let me simplify these equations.\n\nStarting with the first equation:\n\nLeft side: \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nRight side: \$ \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract the left side from the right side to set the equation to zero:\n\n\$ \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 - \\left( \\frac{24}{v - r} + \\frac{28}{v + r} \\right) = 0 \$\n\nSimplify term by term:\n\n\$ \\frac{30 - 24}{v - r} + \\frac{21 - 28}{v + r} - 0.5 = 0 \$\n\nWhich simplifies to:\n\n\$ \\frac{6}{v - r} - \\frac{7}{v + r} - 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\nLeft side: \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nRight side: \$ \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nSubtract left side from right side:\n\n\$ \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 - \\left( \\frac{24}{v - r} + \\frac{28}{v + r} \\right) = 0 \$\n\nSimplify term by term:\n\n\$ \\frac{15 - 24}{v - r} + \\frac{42 - 28}{v + r} + 0.5 = 0 \$\n\nWhich simplifies to:\n\n\$ -\\frac{9}{v - r} + \\frac{14}{v + r} + 0.5 = 0 \$\n\nSo now we have two equations:\n\n1. \$ \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5 \$\n2. \$ -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5 \$\n\nLet me write them as:\n\nEquation (1): \$ \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5 \$\n\nEquation (2): \$ -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5 \$\n\nHmm, these are two equations with two variables, \$ \\frac{1}{v - r} \$ and \$ \\frac{1}{v + r} \$. Let me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\nEquation (1): \$ 6x - 7y = 0.5 \$\n\nEquation (2): \$ -9x + 14y = -0.5 \$\n\nNow, I can solve this system of linear equations for x and y. Let me write them again:\n\n1. \$ 6x - 7y = 0.5 \$\n2. \$ -9x + 14y = -0.5 \$\n\nLet me use the elimination method. If I multiply equation (1) by 2, I get:\n\nEquation (1a): \$ 12x - 14y = 1 \$\n\nNow, equation (2) is \$ -9x + 14y = -0.5 \$\n\nIf I add equation (1a) and equation (2), the y terms will cancel out:\n\n\$ (12x - 14y) + (-9x + 14y) = 1 + (-0.5) \$\n\nSimplify:\n\n\$ 3x = 0.5 \$\n\nTherefore, \$ x = 0.5 / 3 = 1/6 \$\n\nNow, substitute x = 1/6 back into equation (1):\n\n\$ 6*(1/6) -7y = 0.5 \$\n\nSimplify:\n\n\$ 1 - 7y = 0.5 \$\n\nSubtract 1:\n\n\$ -7y = -0.5 \$\n\nDivide:\n\n\$ y = (-0.5)/(-7) = 0.5/7 = 1/14 \$\n\nSo, we have x = 1/6 and y = 1/14. Recall that:\n\n\$ x = \\frac{1}{v - r} = 1/6 \$\n\n\$ y = \\frac{1}{v + r} = 1/14 \$\n\nTherefore, we can write:\n\n\$ v - r = 6 \$\n\n\$ v + r = 14 \$\n\nNow, solving these two equations:\n\nAdd the two equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nWhich simplifies to:\n\n\$ 2v = 20 \$\n\nTherefore, \$ v = 10 \$ km/h\n\nSubstituting back into \$ v - r = 6 \$:\n\n\$ 10 - r = 6 \$\n\nThus, \$ r = 10 - 6 = 4 \$ km/h\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results to ensure there are no mistakes.\n\nFirst, check the first journey: 24 km upstream and 28 km downstream.\n\nUpstream speed: 10 - 4 = 6 km/h. Time upstream: 24 / 6 = 4 hours.\n\nDownstream speed: 10 + 4 = 14 km/h. Time downstream: 28 / 14 = 2 hours.\n\nTotal time: 4 + 2 = 6 hours.\n\nSecond journey: 30 km upstream and 21 km downstream.\n\nTime upstream: 30 / 6 = 5 hours.\n\nTime downstream: 21 / 14 = 1.5 hours.\n\nTotal time: 5 + 1.5 = 6.5 hours.\n\nDifference between the two journeys: 6.5 - 6 = 0.5 hours, which is half an hour. That's correct.\n\nThird journey: 15 km upstream and 42 km downstream.\n\nTime upstream: 15 / 6 = 2.5 hours.\n\nTime downstream: 42 / 14 = 3 hours.\n\nTotal time: 2.5 + 3 = 5.5 hours.\n\nDifference between the first journey (6 hours) and this one: 6 - 5.5 = 0.5 hours, which is half an hour more. That also checks out.\n\nTherefore, the solution seems correct. The speed of the ship in still water is 10 km/h and the speed of the river is 4 km/h.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$ respectively.\n\nWe are given two comparisons involving the time taken for different journeys. The first comparison states that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. The second comparison states that the same journey (24 km upstream and 28 km downstream) took half an hour more than traveling 15 km upstream and 42 km downstream.\n\nWe set up the following equations based on the time differences:\n\n1. For the first comparison:\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n Simplifying this, we get:\n \\[\n \\frac{6}{v - r} - \\frac{7}{v + r} = 0.5\n \\]\n\n2. For the second comparison:\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n Simplifying this, we get:\n \\[\n -\\frac{9}{v - r} + \\frac{14}{v + r} = -0.5\n \\]\n\nIntroducing \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we convert these equations into a system of linear equations:\n1. \$ 6x - 7y = 0.5 \$\n2. \$ -9x + 14y = -0.5 \$\n\nSolving this system, we multiply the first equation by 2:\n1. \$ 12x - 14y = 1 \$\n\nAdding this to the second equation:\n\\[\n3x = 0.5 \\implies x = \\frac{1}{6}\n\\]\n\nSubstituting \$ x = \\frac{1}{6} \$ back into the first equation:\n\\[\n6 \\left( \\frac{1}{6} \\right) - 7y = 0.5 \\implies 1 - 7y = 0.5 \\implies y = \\frac{1}{14}\n\\]\n\nThus, we have:\n\\[\nv - r = 6 \\quad \\text{and} \\quad v + r = 14\n\\]\n\nSolving these equations:\n\\[\n2v = 20 \\implies v = 10 \\quad \\text{and} \\quad r = 10 - 6 = 4\n\\]\n\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h." ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">1</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ { "content": "## Task B-1.3.\n\nA ship traveling along a river has covered $24 \\mathrm{~km}$ upstream and $28 \\mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \\mathrm{~km}$ upstream and $21 \\mathrm{~km}$ downstream, or half an hour more than for traveling $15 \\mathrm{~km}$ upstream and $42 \\mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.\n\nDetermine the speed of the ship in still water and the speed of the river.", "role": "user" }, { "content": "<think>\nOkay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed. \n\nLet me denote the speed of the ship in still water as \$ v \$ (in km/h) and the speed of the river as \$ r \$ (also in km/h). Then, the upstream speed would be \$ v - r \$, and the downstream speed would be \$ v + r \$.\n\nThe problem mentions three different journeys:\n\n1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.\n2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.\n3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.\n\nWait, actually, the problem states: \"For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream...\"\n\nHmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:\n\n- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours\n- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours\n\nTherefore, we can set up two equations based on these time differences.\n\nFirst, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):\n\nTime1 = \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$\n\nFor the second journey (30 km up, 21 km down):\n\nTime2 = \$ \\frac{30}{v - r} + \\frac{21}{v + r} \$\n\nAccording to the problem, Time1 = Time2 - 0.5\n\nSimilarly, for the third journey (15 km up, 42 km down):\n\nTime3 = \$ \\frac{15}{v - r} + \\frac{42}{v + r} \$\n\nAnd here, Time1 = Time3 + 0.5\n\nSo, we have two equations:\n\n1. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n2. \$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5 \$\n\nOkay, so now I need to solve these two equations for \$ v \$ and \$ r \$.\n\nLet me first work on the first equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5 \$\n\nLet me subtract \$ \\frac{24}{v - r} + \\frac{28}{v + r} \$ from both sides, but maybe it's better to bring all terms to the left side:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{30}{v - r} - \\frac{21}{v + r} + 0.5 = 0 \$\n\nSimplify the terms:\n\nFor the upstream terms: \$ 24 - 30 = -6 \$, so \$ \\frac{-6}{v - r} \$\n\nFor the downstream terms: \$ 28 - 21 = 7 \$, so \$ \\frac{7}{v + r} \$\n\nSo, the equation becomes:\n\n\$ \\frac{-6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n\nSimilarly, for the second equation:\n\n\$ \\frac{24}{v - r} + \\frac{28}{v + r} - \\frac{15}{v - r} - \\frac{42}{v + r} - 0.5 = 0 \$\n\nSimplify:\n\nUpstream terms: \$ 24 - 15 = 9 \$, so \$ \\frac{9}{v - r} \$\n\nDownstream terms: \$ 28 - 42 = -14 \$, so \$ \\frac{-14}{v + r} \$\n\nThus:\n\n\$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nNow, we have two equations:\n\n1. \$ -\\frac{6}{v - r} + \\frac{7}{v + r} + 0.5 = 0 \$\n2. \$ \\frac{9}{v - r} - \\frac{14}{v + r} - 0.5 = 0 \$\n\nLet me denote \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$. Then, the equations become:\n\n1. \$ -6x + 7y + 0.5 = 0 \$ ⇒ \$ -6x + 7y = -0.5 \$\n2. \$ 9x - 14y - 0.5 = 0 \$ ⇒ \$ 9x - 14y = 0.5 \$\n\nNow, we have a system of linear equations:\n\nEquation (1): \$ -6x + 7y = -0.5 \$\n\nEquation (2): \$ 9x - 14y = 0.5 \$\n\nLet me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:\n\nEquation (1) multiplied by 2: \$ -12x + 14y = -1 \$\n\nNow, add this to Equation (2):\n\n\$ (-12x + 14y) + (9x -14y) = -1 + 0.5 \$\n\nSimplify:\n\n\$ -3x = -0.5 \$\n\nThus, \$ x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \$\n\nBut exact value is \$ x = 1/6 \$\n\nNow, substitute \$ x = 1/6 \$ back into Equation (1):\n\n\$ -6*(1/6) + 7y = -0.5 \$\n\nSimplify:\n\n\$ -1 + 7y = -0.5 \$\n\nAdd 1 to both sides:\n\n7y = 0.5\n\nThus, \$ y = 0.5/7 = 1/14 ≈ 0.0714 \$\n\nSo, \$ x = 1/6 \$ and \$ y = 1/14 \$\n\nRecall that \$ x = \\frac{1}{v - r} = 1/6 \$ ⇒ \$ v - r = 6 \$\n\nSimilarly, \$ y = \\frac{1}{v + r} = 1/14 \$ ⇒ \$ v + r = 14 \$\n\nTherefore, we have two equations:\n\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nAdding these equations:\n\n\$ (v - r) + (v + r) = 6 + 14 \$\n\nSimplify:\n\n\$ 2v = 20 \$ ⇒ \$ v = 10 \$\n\nSubstituting back into \$ v + r = 14 \$:\n\n\$ 10 + r = 14 \$ ⇒ \$ r = 4 \$\n\nSo, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.\n\nLet me verify these results by plugging them back into the original time equations.\n\nFirst journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.\n\nSecond journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.\n\nThird journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.\n\nTherefore, the solution is correct.\n\n**Final Answer**\nThe speed of the ship in still water is \\boxed{10} km/h and the speed of the river is \\boxed{4} km/h.\n</think>\n\nLet \$ v \$ be the speed of the ship in still water (in km/h) and \$ r \$ be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \$ v - r \$ and \$ v + r \$, respectively.\n\nWe are given three journeys with the following time relationships:\n1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.\n2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.\n\nWe set up the equations for the times taken for each journey:\n1. For the first journey (24 km upstream, 28 km downstream):\n \\[\n \\frac{24}{v - r} + \\frac{28}{v + r}\n \\]\n2. For the second journey (30 km upstream, 21 km downstream):\n \\[\n \\frac{30}{v - r} + \\frac{21}{v + r}\n \\]\n3. For the third journey (15 km upstream, 42 km downstream):\n \\[\n \\frac{15}{v - r} + \\frac{42}{v + r}\n \\]\n\nThe time relationships give us the following equations:\n1. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{30}{v - r} + \\frac{21}{v + r} - 0.5\n \\]\n2. \\[\n \\frac{24}{v - r} + \\frac{28}{v + r} = \\frac{15}{v - r} + \\frac{42}{v + r} + 0.5\n \\]\n\nIntroducing variables \$ x = \\frac{1}{v - r} \$ and \$ y = \\frac{1}{v + r} \$, we transform the equations into:\n1. \\[\n -6x + 7y = -0.5\n \\]\n2. \\[\n 9x - 14y = 0.5\n \\]\n\nSolving this system of linear equations:\n1. Multiply the first equation by 2:\n \\[\n -12x + 14y = -1\n \\]\n2. Add the modified first equation to the second equation:\n \\[\n -3x = -0.5 \\implies x = \\frac{1}{6}\n \\]\n3. Substitute \$ x = \\frac{1}{6} \$ into the first equation:\n \\[\n -6 \\left( \\frac{1}{6} \\right) + 7y = -0.5 \\implies -1 + 7y = -0.5 \\implies 7y = 0.5 \\implies y = \\frac{1}{14}\n \\]\n\nConverting back to \$ v \$ and \$ r \$:\n1. \$ v - r = 6 \$\n2. \$ v + r = 14 \$\n\nSolving these equations:\n1. Adding the two equations:\n \\[\n 2v = 20 \\implies v = 10\n \\]\n2. Substituting \$ v = 10 \$ into \$ v + r = 14 \$:\n \\[\n 10 + r = 14 \\implies r = 4\n \\]\n\nThe speed of the ship in still water is \$\\boxed{10}\$ km/h and the speed of the river is \$\\boxed{4}\$ km/h.", "role": "assistant" } ]</div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="1"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel.</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">3. (6 points) A construction company builds a tunnel. When $\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \%$ and reducing the daily working hours to $80 \%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\qquad$ 180 days. 【Solution】Solution: $\left(1-\frac{1}{3}\right) \div[(1+20 \%) \times 80 \%]$ $$ \begin{array}{l} =\frac{2}{3} \div[120 \% \times 80 \%], \\ =\frac{2}{3} \div \frac{24}{25} \\ =\frac{25}{36} ; \\ 185 \div\left(\frac{1}{3}+\frac{25}{36}\right) \\ =185 \div \frac{37}{36}, \\ =180 \text { (days). } \end{array} $$ Answer: If the original speed was maintained, it would take 180 days. Therefore, the answer is: 180.</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">180</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Algebra</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">1d57f4b3-c516-5522-91cc-a60dd91f6f66</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ "<think>\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\n\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\n\nWait, the problem says \"reduced the working hours to 80% of the original.\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\n\nFirst, let's define some variables. Let's let the original construction speed be \"v\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\n\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\n\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\n\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\n\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\n\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\n\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\n\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\n\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\n\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\n\nBut total time is 185 days, so:\n\nTime for Phase 1 + Time for Phase 2 = 185\n\nWhich is:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nWe can solve for R, then T = 1/R is the original total time.\n\nLet me compute that.\n\nFirst, let's write the equation:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nCombine the terms:\n\n[1/3 + (2/3)/0.96] / R = 185\n\nCalculate the numerator:\n\n1/3 + (2/3)/0.96\n\nFirst compute (2/3)/0.96:\n\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\n\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\n\nSo numerator ≈ 1.0277\n\nTherefore, 1.0277 / R = 185\n\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\n\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\n\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\n\nLet me compute 1/3 + (2/3)/0.96 exactly.\n\nFirst, (2/3)/0.96:\n\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\n\nThus, (2/3)*(25/24) = (50/72) = 25/36.\n\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\n\nTherefore, numerator is 37/36.\n\nSo the equation is (37/36)/R = 185\n\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\n\nTherefore, T = 1/R = 180 days.\n\nSo the original time would be 180 days.\n\nWait, so the answer is 180 days? Let me confirm.\n\nIf original time is 180 days, then original rate is 1/180 per day.\n\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\n\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\n\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\n\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\n\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\n\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\n\nTherefore, if they had not used the new equipment, it would have taken 180 days.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original number of days required to complete the tunnel. The original construction speed is \$ \\frac{1}{T} \$ tunnels per day. \n\nWhen \$ \\frac{1}{3} \$ of the tunnel was completed at the original speed, the time taken for this phase is:\n\\[\n\\frac{1/3}{1/T} = \\frac{T}{3} \\text{ days}\n\\]\n\nFor the remaining \$ \\frac{2}{3} \$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \$ 1.2 \\times \\frac{1}{T} \$ per hour, but due to reduced working hours, the effective daily rate becomes:\n\\[\n1.2 \\times \\frac{1}{T} \\times 0.8 = \\frac{0.96}{T} \\text{ tunnels per day}\n\\]\n\nThe time required to complete the remaining \$ \\frac{2}{3} \$ of the tunnel at the new rate is:\n\\[\n\\frac{2/3}{0.96/T} = \\frac{2T}{3 \\times 0.96} = \\frac{2T}{2.88} = \\frac{25T}{36} \\text{ days}\n\\]\n\nThe total time taken with the new equipment is 185 days:\n\\[\n\\frac{T}{3} + \\frac{25T}{36} = 185\n\\]\n\nCombining the terms:\n\\[\n\\frac{12T}{36} + \\frac{25T}{36} = \\frac{37T}{36} = 185\n\\]\n\nSolving for \$ T \$:\n\\[\nT = 185 \\times \\frac{36}{37} = 180\n\\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel.", "<think>\nAlright, let's try to solve this problem step by step. So, we have a construction company building a tunnel. They completed the first 1/3 of the tunnel at the original speed, then started using new equipment which increased their speed by 20% and reduced their working hours to 80% of the original. The total time taken was 185 days. We need to find out how many days it would have taken if they didn't use the new equipment and continued at the original speed.\n\nOkay, let's break this down. Let's denote the original speed as S (maybe in terms of tunnel length per day), and the original working hours per day as H. Then, with the new equipment, their speed becomes S + 20% of S, which is 1.2S. Their working hours become 80% of H, which is 0.8H.\n\nWait, but how does working hours per day affect the total days? Hmm. If they reduce the working hours per day, then even if their speed is higher, the amount they can do each day would be speed multiplied by hours. So maybe we need to consider the daily progress before and after the equipment change.\n\nLet me think. Let's assume that originally, without any new equipment, the total tunnel length is L. Then, the total work is L. The original speed is S, so the time to complete the tunnel would be T = L / S days. But they might have a certain number of hours per day, so maybe S is already accounting for hours per day? Wait, the problem mentions \"reduced the working hours to 80% of the original\". So working hours per day is a factor here.\n\nWait, perhaps it's better to model this in terms of man-hours or something? Let's see. Let me try to structure it.\n\nLet me denote:\n\n- Original speed: S (tunnel length per hour)\n\n- Original working hours per day: H\n\nSo originally, the amount of tunnel they can build in one day is S * H.\n\nAfter the new equipment:\n\n- Speed increases by 20%, so new speed is 1.2S\n\n- Working hours per day decrease to 80% of original, so new working hours per day is 0.8H\n\nTherefore, the new daily progress is 1.2S * 0.8H = 0.96SH\n\nWait, that's interesting. So even though speed increased by 20%, the working hours decreased to 80%, so the total daily progress is 0.96 times the original daily progress. So actually, the daily progress decreased by 4%?\n\nBut that seems counterintuitive. Let me check that math again.\n\nOriginal daily progress: S * H\n\nNew daily progress: (1.2S) * (0.8H) = 1.2 * 0.8 * S * H = 0.96SH. Yes, so 0.96 times original daily progress. So actually, despite increasing speed by 20%, the reduction in working hours led to a 4% decrease in daily progress.\n\nHmm, so their daily rate of construction actually decreased slightly when they changed the equipment? But they still managed to complete the tunnel in 185 days. But if they had continued at the original speed and hours, how long would it have taken?\n\nWait, maybe my approach is missing something. Let's see. The problem says \"increased the construction speed by 20% and reduced the working hours to 80% of the original\". So speed per hour is increased by 20%, but they are working fewer hours each day.\n\nAlternatively, maybe \"construction speed\" is already in terms of tunnel length per day? Then, increasing the speed by 20% would mean 1.2 times the original daily speed, but then they reduced the working hours to 80%. Wait, that might be a different interpretation.\n\nWait, the problem is a bit ambiguous. The construction speed could be per hour or per day. The problem says \"increased the construction speed by 20%\". If the original speed was, say, x tunnel length per hour, then a 20% increase would make it 1.2x per hour. Then, if they reduce working hours to 80% of original, the daily progress becomes 1.2x * 0.8H = 0.96xH, which is again 96% of the original daily progress. So the same as before.\n\nAlternatively, if the original speed was given as tunnel length per day, and then they increased that speed by 20%, but then reduced the working hours. Hmm. Wait, the problem says \"increased the construction speed by 20%\". So \"construction speed\" is probably the rate at which they construct, which could be per hour. Then, when they reduce the working hours, their daily progress is (speed per hour)*(hours per day).\n\nSo in any case, the daily progress becomes 0.96 times the original daily progress. So even though they work faster per hour, they work fewer hours each day, resulting in a slight decrease in daily progress.\n\nBut according to the problem, after they started using the new equipment, they managed to complete the tunnel in a total of 185 days. But how does that compare to the original plan?\n\nWait, let's structure the problem.\n\nFirst, they build 1/3 of the tunnel at the original speed. Then, they build the remaining 2/3 with the new equipment. The total time is 185 days. If they hadn't used the new equipment, they would have continued at the original speed for the entire tunnel, and that hypothetical total time is what we need to find.\n\nSo let's denote:\n\nLet T be the original total time needed to build the tunnel without any new equipment. Then, the original daily progress is L / T, where L is the total length.\n\nWait, maybe that's a better way. Let me assume the total length of the tunnel is 1 unit for simplicity. Then, original total time is T days. So original daily progress is 1/T per day.\n\nBut wait, when they start using the new equipment after 1/3 of the tunnel is built, they change their daily progress. Let's figure out the time taken for the first 1/3 and the time taken for the remaining 2/3.\n\nBut how is the daily progress affected?\n\nIf originally, the entire tunnel takes T days, then the original daily progress is 1/T per day. So the first 1/3 of the tunnel would take (1/3) / (1/T) = T/3 days.\n\nAfter that, they switch to the new equipment. With the new equipment, their daily progress is 0.96 * (1/T) per day, as we calculated before? Wait, but no. Wait, if the original daily progress is 1/T per day (since they complete 1 tunnel in T days), then with the new equipment, their daily progress is 0.96*(1/T). Therefore, the remaining 2/3 of the tunnel would take (2/3) / (0.96/T) = (2/3)*(T/0.96) = (2T)/(3*0.96) days.\n\nTherefore, the total time is T/3 + (2T)/(3*0.96) = 185 days.\n\nSo solving for T:\n\nT/3 + (2T)/(3*0.96) = 185\n\nLet me compute this equation.\n\nFirst, factor out T/3:\n\nT/3 [1 + 2/0.96] = 185\n\nCompute 2/0.96: 2 divided by 0.96 is equal to 2.083333...\n\nSo, 1 + 2.083333... = 3.083333...\n\nTherefore, T/3 * 3.083333... = 185\n\nMultiply both sides by 3:\n\nT * 3.083333... = 555\n\nTherefore, T = 555 / 3.083333...\n\nCompute 3.083333... as a fraction. 3.083333... is 3 + 0.083333..., and 0.083333... is 1/12. So 3 + 1/12 = 37/12.\n\nTherefore, T = 555 / (37/12) = 555 * (12/37) = (555/37)*12\n\nCompute 555 divided by 37:\n\n37*15 = 555, so 555/37 = 15\n\nTherefore, T = 15*12 = 180 days.\n\nWait, so the original total time would have been 180 days, and according to the problem, with the new equipment, it took 185 days. That seems like using the new equipment actually made them slower? Because 185 is more than 180. But the problem states that \"as a result, it took a total of 185 days\". Wait, maybe I messed up the daily progress.\n\nWait, hold on. If the original total time is T days, then when they switch to the new equipment after 1/3 of the tunnel is done, which took T/3 days, and then the remaining 2/3 is done at a slower daily rate (0.96 of original daily rate), which would take longer, so total time becomes longer than T? But that contradicts the problem statement, which probably implies that using the new equipment should have made them faster, but maybe not? Wait, maybe not. Wait, let's check.\n\nWait, according to the calculation, when they use the new equipment, their daily progress is 96% of the original, so they actually progress slower each day, so the remaining 2/3 takes longer. So even though they increased speed by 20%, the reduction in working hours caused their daily progress to decrease. Therefore, the total time increased from 180 days to 185 days. But the question is, if they had NOT used the new equipment, how long would it have taken? The answer is 180 days.\n\nBut let's check again if my reasoning is correct.\n\nLet me verify with numbers. Suppose the original total time is 180 days.\n\nFirst 1/3 of the tunnel: 180 / 3 = 60 days.\n\nRemaining 2/3 at original speed: 180 - 60 = 120 days.\n\nBut with the new equipment, daily progress is 0.96 of original. So time taken for remaining 2/3 is 120 / 0.96 = 125 days.\n\nTotal time: 60 + 125 = 185 days. Which matches the problem's statement.\n\nSo that makes sense. Therefore, the original total time is 180 days. Therefore, the answer is 180.\n\nBut let me think again. The problem says \"increased the construction speed by 20% and reduced the working hours to 80% of the original\". So construction speed (per hour) increased by 20%, but working hours per day decreased to 80%. So the daily progress is 1.2 * 0.8 = 0.96 times original. So 96% of original daily progress. Therefore, the remaining 2/3 would take (2/3)/0.96 of the original time. Original time for 2/3 is (2/3)*T, so new time is (2/3)*T /0.96 = (2T)/(3*0.96). Then total time is T/3 + (2T)/(3*0.96) = 185.\n\nSolving for T gives 180. Therefore, the answer is 180 days. That seems correct.\n\nAlternatively, maybe my initial interpretation of speed is wrong. Let's try another approach.\n\nSuppose the original speed is S (length per day). Then, the total length L = S * T, where T is the original total time.\n\nThey complete 1/3 of the tunnel at speed S, which takes (L/3)/S = T/3 days.\n\nThen, they use new equipment, which increases speed by 20%, so new speed is 1.2S. But working hours are reduced to 80% of original. Wait, if the original speed S is length per day, which is dependent on working hours per day. So if they reduce working hours per day to 80%, but speed per hour increases by 20%, then the new speed in terms of length per day would be 1.2 * original speed per hour * 0.8 * original hours per day.\n\nIf original speed per hour is s, then original speed per day is s * H, where H is original hours per day. Then, new speed per day is (1.2s) * (0.8H) = 0.96sH = 0.96S. Therefore, new speed per day is 0.96S.\n\nTherefore, remaining 2/3 of the tunnel would take (2L/3)/(0.96S) = (2/3)*(L/S)/0.96 = (2/3)*T /0.96 = (2T)/(3*0.96) days.\n\nSo total time is T/3 + (2T)/(3*0.96) = 185.\n\nWhich leads to T = 180. So same result.\n\nTherefore, regardless of whether we model speed as per day or per hour, we get the same result. Therefore, the original total time would have been 180 days.\n\nTherefore, the answer is 180, boxed.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original total time needed to complete the tunnel without using the new equipment. The tunnel is divided into two parts: the first \$ \\frac{1}{3} \$ completed at the original speed, and the remaining \$ \\frac{2}{3} \$ completed with the new equipment.\n\n1. **Time to complete the first \$ \\frac{1}{3} \$ of the tunnel:**\n \\[\n \\frac{T}{3} \\text{ days}\n \\]\n\n2. **Effect of new equipment:**\n - Construction speed increases by 20%, so the new speed is \$ 1.2 \$ times the original speed.\n - Working hours are reduced to 80% of the original, so the new working hours are \$ 0.8 \$ times the original.\n\n The new daily progress is \$ 1.2 \\times 0.8 = 0.96 \$ times the original daily progress. Therefore, the time to complete the remaining \$ \\frac{2}{3} \$ of the tunnel with the new equipment is:\n \\[\n \\frac{\\frac{2}{3}T}{0.96} = \\frac{2T}{3 \\times 0.96}\n \\]\n\n3. **Total time with new equipment:**\n \\[\n \\frac{T}{3} + \\frac{2T}{3 \\times 0.96} = 185\n \\]\n\n4. **Solving for \$ T \$:**\n \\[\n \\frac{T}{3} + \\frac{2T}{2.88} = 185\n \\]\n \\[\n \\frac{T}{3} + \\frac{5T}{7.2} = 185\n \\]\n Converting to a common denominator:\n \\[\n \\frac{2.4T}{7.2} + \\frac{5T}{7.2} = 185\n \\]\n \\[\n \\frac{7.4T}{7.2} = 185\n \\]\n \\[\n T = 185 \\times \\frac{7.2}{7.4}\n \\]\n \\[\n T = 180\n \\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel." ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ { "content": "3. (6 points) A construction company was building a tunnel. When $\\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \\%$ and reduced the working hours to $80 \\%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\\qquad$ days to complete the tunnel.", "role": "user" }, { "content": "<think>\nOkay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.\n\nHmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.\n\nWait, the problem says \"reduced the working hours to 80% of the original.\" So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.\n\nFirst, let's define some variables. Let's let the original construction speed be \"v\" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.\n\nBut when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.\n\nWait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.\n\nFirst, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.\n\nLet me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.\n\nIn Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.\n\nIn Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?\n\nWait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?\n\nWait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.\n\nWait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.\n\nBut total time is 185 days, so:\n\nTime for Phase 1 + Time for Phase 2 = 185\n\nWhich is:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nWe can solve for R, then T = 1/R is the original total time.\n\nLet me compute that.\n\nFirst, let's write the equation:\n\n1/(3R) + (2/3)/(0.96R) = 185\n\nCombine the terms:\n\n[1/3 + (2/3)/0.96] / R = 185\n\nCalculate the numerator:\n\n1/3 + (2/3)/0.96\n\nFirst compute (2/3)/0.96:\n\n2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944\n\nThen add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277\n\nSo numerator ≈ 1.0277\n\nTherefore, 1.0277 / R = 185\n\nSo R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day\n\nTherefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days\n\nWait, so is T 180 days? Let me check the exact calculation instead of approximating.\n\nLet me compute 1/3 + (2/3)/0.96 exactly.\n\nFirst, (2/3)/0.96:\n\n0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.\n\nThus, (2/3)*(25/24) = (50/72) = 25/36.\n\nThen 1/3 is 12/36, so 12/36 + 25/36 = 37/36.\n\nTherefore, numerator is 37/36.\n\nSo the equation is (37/36)/R = 185\n\nHence, R = 37/(36*185) = 37/(6660) = 1/180 per day.\n\nTherefore, T = 1/R = 180 days.\n\nSo the original time would be 180 days.\n\nWait, so the answer is 180 days? Let me confirm.\n\nIf original time is 180 days, then original rate is 1/180 per day.\n\nPhase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.\n\nPhase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.\n\nTime for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.\n\nWait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.\n\nSo (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.\n\nWait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.\n\nTherefore, if they had not used the new equipment, it would have taken 180 days.\n\n**Final Answer**\n\\boxed{180}\n</think>\n\nLet \$ T \$ be the original number of days required to complete the tunnel. The original construction speed is \$ \\frac{1}{T} \$ tunnels per day. \n\nWhen \$ \\frac{1}{3} \$ of the tunnel was completed at the original speed, the time taken for this phase is:\n\\[\n\\frac{1/3}{1/T} = \\frac{T}{3} \\text{ days}\n\\]\n\nFor the remaining \$ \\frac{2}{3} \$ of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \$ 1.2 \\times \\frac{1}{T} \$ per hour, but due to reduced working hours, the effective daily rate becomes:\n\\[\n1.2 \\times \\frac{1}{T} \\times 0.8 = \\frac{0.96}{T} \\text{ tunnels per day}\n\\]\n\nThe time required to complete the remaining \$ \\frac{2}{3} \$ of the tunnel at the new rate is:\n\\[\n\\frac{2/3}{0.96/T} = \\frac{2T}{3 \\times 0.96} = \\frac{2T}{2.88} = \\frac{25T}{36} \\text{ days}\n\\]\n\nThe total time taken with the new equipment is 185 days:\n\\[\n\\frac{T}{3} + \\frac{25T}{36} = 185\n\\]\n\nCombining the terms:\n\\[\n\\frac{12T}{36} + \\frac{25T}{36} = \\frac{37T}{36} = 185\n\\]\n\nSolving for \$ T \$:\n\\[\nT = 185 \\times \\frac{36}{37} = 180\n\\]\n\nThus, if they had not used the new equipment and continued at the original speed, it would have taken \$\\boxed{180}\$ days to complete the tunnel.", "role": "assistant" } ]</div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="2"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"Prove that number $1$ can be represented as a sum of a finite number $n$ of real numbers, less than<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"1. **Restate the problem in a more manageable form:**\n We need to prove that the number \$1\$ <span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto"> 8 </div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Number Theory</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">aops_forum</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">1cd82e3d-2e7a-5eca-9571-62f5669824c4</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so the problem is to prove that the number 1 can be represented as a sum of a finit<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">1</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"Prove that number $1$ can be represented as a sum of a finite number $n$ of real number<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="3"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"4. Given the three sides of an obtuse triangle are 3, $4, x$, then the range of values for $x$ is (<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">$D$</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">D</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Geometry</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">MCQ</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">cn_contest</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">ea24dff6-9e4d-5c55-9ec0-c4d55424997a</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so I have this problem here where I need to find the range of values for x, given t<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"4. Given the three sides of an obtuse triangle are 3, $4, x$, then the range of values <span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="4"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"1. Solve the equation: $\\frac{8 x+13}{3}=\\frac{19-12 x}{2}-\\left(16 x-\\frac{7-4 x}{6}\\right)$.<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"1. $\\frac{8 x+13}{3}=\\frac{19-12 x}{2}-\\left(16 x-\\frac{7-4 x}{6}\\right)$\n\n$\\frac{8 x+13}{3<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">\frac{1}{4}</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Algebra</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">a583d65c-9b4c-56ce-8263-47edbc22baa3</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so I need to solve this equation: (8x + 13)/3 = (19 - 12x)/2 - (16x - (7 - 4x)/6). <span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"1. Solve the equation: $\\frac{8 x+13}{3}=\\frac{19-12 x}{2}-\\left(16 x-\\frac{7-4 x}{<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="5"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"A right-angled triangle has side lengths that are integers. What could be the last digit of the are<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to t<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">0</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Geometry</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">79e837be-2801-5a6a-bc7a-0acee3cdd5fd</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so I need to figure out the possible last digits of the area of a right-angled tria<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"A right-angled triangle has side lengths that are integers. What could be the last digi<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="6"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"Let $a_n\\ (n\\geq 1)$ be the value for which $\\int_x^{2x} e^{-t^n}dt\\ (x\\geq 0)$ is maximal. Fi<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"1. Define the integral \$ I_n(x) = \\int_x^{2x} e^{-t^n} \\, dt \$. We need to find the value of <span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">-\ln 2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Calculus</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">aops_forum</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">167ecbbb-4270-534f-afa5-75d01d968ab5</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true, true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so I need to find the limit as n approaches infinity of the natural logarithm of a_<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true, true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">4</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"Let $a_n\\ (n\\geq 1)$ be the value for which $\\int_x^{2x} e^{-t^n}dt\\ (x\\geq 0)$ is<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="7"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"11. Given that the internal angles $A, B, C$ of $\\triangle ABC$ have opposite sides $a, b, c$ resp<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"(1) $\\sqrt{3} \\sin B \\cos \\frac{A+B}{2}=\\sin C \\sin B \\Rightarrow \\sqrt{3} \\sin \\frac{C}{<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">\sinA=\frac{1}{2}or\sinA=1</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Geometry</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">84f224e8-719b-5b42-a410-27816014bdd3</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, so I have this triangle ABC, with angles A, B, C opposite sides a, b, c respectivel<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ "stop", "stop" ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">2</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"11. Given that the internal angles $A, B, C$ of $\\triangle ABC$ have opposite sides $a<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="8"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digits of the nu<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"## Solution.\n\nIf between every two digits of the number 14641 we write 2013 zeros, we get a numbe<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">x_{1,2}=\(10^{2014}+1),\quadx_{3,4}=\(10^{2014}+1)\cdoti</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Algebra</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">olympiads</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">f769fc87-fed5-5b73-8403-372cdd409800</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, true ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, let's see. I need to find all complex solutions to the equation x⁴ = n, where n i<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ false, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">1</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digi<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td> </tr><tr class="group cursor-pointer space-x-4 divide-x border-b outline-offset-[-2px] odd:bg-gray-50 hover:bg-gray-100 dark:odd:bg-gray-925 dark:hover:bg-gray-850 last:border-none" tabindex="0" data-row-idx="9"><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"6. As shown in Figure 2, let $P$ be a point inside the equilateral $\\triangle ABC$ with side lengt<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">"$$\n6.11 \\sqrt{3} \\text {. }\n$$\n\nFor the equilateral $\\triangle ABC$ with a height of $12 \\t<span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">11 \sqrt{3}</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">Geometry</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">math-word-problem</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">cn_contest</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="" dir="auto">2a4c0374-d9ea-538e-8e0f-0811359d6cf6</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">["<think>\nOkay, let me try to solve this geometry problem. So, we have an equilateral triangle ABC <span class="text-orange-500">(...TRUNCATED)</span></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[ true, false ]</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto"><div class="text-right text-gray-400">null</div></div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-right" dir="auto">1</div></div> </td><td class="min-w-fit max-w-sm break-words p-2 "><div class="line-clamp-2 "><div class="text-left" dir="auto">[{"content":"6. 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data-props="{"titleTree":[{"id":"dataset-description","label":"Dataset description","children":[],"isValid":true,"title":"Dataset description"},{"id":"dataset-curation","label":"Dataset curation","children":[],"isValid":true,"title":"Dataset curation"},{"id":"license","label":"License","children":[],"isValid":true,"title":"License"}],"classNames":"top-6"}"> <div class="absolute -left-12 bottom-0 top-0 z-10 top-6"><div class="sticky top-4 flex"><div class="h-7 pt-[0.175rem]"> <span class="peer" tabindex="0"><button class="select-none hover:cursor-pointer"><svg width="1em" height="1em" viewBox="0 0 10 10" class="text-lg opacity-80 hover:opacity-100" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" preserveAspectRatio="xMidYMid meet" fill="currentColor"><path fill-rule="evenodd" clip-rule="evenodd" d="M1.65039 2.9999C1.65039 2.8066 1.80709 2.6499 2.00039 2.6499H8.00039C8.19369 2.6499 8.35039 2.8066 8.35039 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md:group-hover:opacity-100 md:right-full" id="openr1-math-220k"> <span class="header-link"><svg viewBox="0 0 256 256" preserveAspectRatio="xMidYMid meet" height="1em" width="1em" role="img" aria-hidden="true" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4"><path fill="currentColor" d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z"></path></svg></span> </a> <span> OpenR1-Math-220k </span> </h1> <h2 class="relative group flex items-center"> <a rel="nofollow" href="#dataset-description" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" id="dataset-description"> <span class="header-link"><svg viewBox="0 0 256 256" preserveAspectRatio="xMidYMid meet" height="1em" width="1em" role="img" aria-hidden="true" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4"><path fill="currentColor" d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z"></path></svg></span> </a> <span> Dataset description </span> </h2> <p>OpenR1-Math-220k is a large-scale dataset for mathematical reasoning. It consists of 220k math problems with two to four reasoning traces generated by <a rel="nofollow" href="https://huggingface.co/deepseek-ai/DeepSeek-R1">DeepSeek R1</a> for problems from NuminaMath 1.5. The traces were verified using <a rel="nofollow" href="https://github.com/huggingface/Math-Verify">Math Verify</a> for most samples and <a rel="nofollow" href="https://huggingface.co/meta-llama/Llama-3.3-70B-Instruct">Llama-3.3-70B-Instruct</a> as a judge for 12% of the samples, and each problem contains at least one reasoning trace with a correct answer.</p> <p>The dataset consists of two splits: </p> <ul> <li><code>default</code> with 94k problems and that achieves the best performance after SFT.</li> <li><code>extended</code> with 131k samples where we add data sources like <code>cn_k12</code>. This provides more reasoning traces, but we found that the performance after SFT to be lower than the <code>default</code> subset, likely because the questions from <code>cn_k12</code> are less difficult than other sources.</li> </ul> <p>You can load the dataset as follows:</p> <pre><code class="language-python"><span class="hljs-keyword">from</span> datasets <span class="hljs-keyword">import</span> load_dataset ds = load_dataset(<span class="hljs-string">"open-r1/OpenR1-Math-220k"</span>, <span class="hljs-string">"default"</span>) </code></pre> <h2 class="relative group flex items-center"> <a rel="nofollow" href="#dataset-curation" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" id="dataset-curation"> <span class="header-link"><svg viewBox="0 0 256 256" preserveAspectRatio="xMidYMid meet" height="1em" width="1em" role="img" aria-hidden="true" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4"><path fill="currentColor" d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z"></path></svg></span> </a> <span> Dataset curation </span> </h2> <p>To build OpenR1-Math-220k, we prompt <a rel="nofollow" href="https://huggingface.co/deepseek-ai/DeepSeek-R1">DeepSeek R1</a> model to generate solutions for 400k problems from <a rel="nofollow" href="https://huggingface.co/datasets/AI-MO/NuminaMath-1.5">NuminaMath 1.5</a> using <a rel="nofollow" href="https://github.com/sgl-project/sglang">SGLang</a>, the generation code is available <a rel="nofollow" href="https://github.com/huggingface/open-r1/tree/main/slurm">here</a>. We follow the model card’s recommended generation parameters and prepend the following instruction to the user prompt:</p> <p><code>"Please reason step by step, and put your final answer within \boxed{}."</code></p> <p>We set a 16k token limit per generation, as our analysis showed that only 75% of problems could be solved in under 8k tokens, and most of the remaining problems required the full 16k tokens. We were able to generate 25 solutions per hour per H100, enabling us to generate 300k problem solutions per day on 512 H100s.</p> <p>We generate two solutions per problem—and in some cases, four—to provide flexibility in filtering and training. This approach allows for rejection sampling, similar to DeepSeek R1’s methodology, and also makes the dataset suitable for preference optimisation methods like DPO.</p> <h2 class="relative group flex items-center"> <a rel="nofollow" href="#license" class="block pr-1.5 text-lg md:absolute md:p-1.5 md:opacity-0 md:group-hover:opacity-100 md:right-full" id="license"> <span class="header-link"><svg viewBox="0 0 256 256" preserveAspectRatio="xMidYMid meet" height="1em" width="1em" role="img" aria-hidden="true" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" class="text-gray-500 hover:text-black dark:hover:text-gray-200 w-4"><path fill="currentColor" d="M167.594 88.393a8.001 8.001 0 0 1 0 11.314l-67.882 67.882a8 8 0 1 1-11.314-11.315l67.882-67.881a8.003 8.003 0 0 1 11.314 0zm-28.287 84.86l-28.284 28.284a40 40 0 0 1-56.567-56.567l28.284-28.284a8 8 0 0 0-11.315-11.315l-28.284 28.284a56 56 0 0 0 79.196 79.197l28.285-28.285a8 8 0 1 0-11.315-11.314zM212.852 43.14a56.002 56.002 0 0 0-79.196 0l-28.284 28.284a8 8 0 1 0 11.314 11.314l28.284-28.284a40 40 0 0 1 56.568 56.567l-28.285 28.285a8 8 0 0 0 11.315 11.314l28.284-28.284a56.065 56.065 0 0 0 0-79.196z"></path></svg></span> </a> <span> License </span> </h2> <p>The dataset is licensed under Apache 2.0</p> </div> </div></section> <section class="pt-6 border-gray-100 md:pb-24 md:pl-6 md:w-64 lg:w-80 xl:w-96 flex-none order-first md:order-none md:border-l !pt-3 md:!pt-6"><dl class="flex items-baseline justify-between"><dt class="text-sm text-gray-500">Downloads last month</dt><div class="mx-4 flex-1 border-b border-dotted"></div><dd class="font-semibold">2,426</dd></dl> <div class="divider-column-vertical"></div> <div class="grid grid-cols-2 gap-x-2 md:flex md:flex-row md:flex-wrap"><div class="SVELTE_HYDRATER contents" data-target="DatasetAndModelActionsDropdown" 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xl:flex-none","libraries":[{"language":"python","library":"datasets","function":"load_dataset","loading_codes":[{"config_name":"default","arguments":{"config_name":"default"},"code":"from datasets import load_dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nds = load_dataset(\"open-r1/OpenR1-Math-220k\", \"default\")"},{"config_name":"all","arguments":{"config_name":"all"},"code":"from datasets import load_dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nds = load_dataset(\"open-r1/OpenR1-Math-220k\", \"all\")"},{"config_name":"extended","arguments":{"config_name":"extended"},"code":"from datasets import load_dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nds = load_dataset(\"open-r1/OpenR1-Math-220k\", \"extended\")"}]},{"language":"python","library":"dask","function":"dd.read_parquet","loading_codes":[{"config_name":"all","arguments":{"splits":{"train":"all/train-*.parquet"}},"code":"import dask.dataframe as dd\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = dd.read_parquet(\"hf://datasets/open-r1/OpenR1-Math-220k/all/train-*.parquet\")"},{"config_name":"default","arguments":{"splits":{"train":"data/train-*.parquet"}},"code":"import dask.dataframe as dd\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = dd.read_parquet(\"hf://datasets/open-r1/OpenR1-Math-220k/data/train-*.parquet\")"},{"config_name":"extended","arguments":{"splits":{"train":"extended/train-*.parquet"}},"code":"import dask.dataframe as dd\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = dd.read_parquet(\"hf://datasets/open-r1/OpenR1-Math-220k/extended/train-*.parquet\")"}]},{"language":"python","library":"mlcroissant","function":"Dataset","loading_codes":[{"config_name":"default","arguments":{"record_set":"default","partial":false},"code":"import requests\nfrom huggingface_hub.file_download import build_hf_headers\nfrom mlcroissant import Dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nheaders = build_hf_headers() # handles authentication\njsonld = requests.get(\"https://huggingface.co/api/datasets/open-r1/OpenR1-Math-220k/croissant\", headers=headers).json()\nds = Dataset(jsonld=jsonld)\nrecords = ds.records(\"default\")"},{"config_name":"all","arguments":{"record_set":"all","partial":false},"code":"import requests\nfrom huggingface_hub.file_download import build_hf_headers\nfrom mlcroissant import Dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nheaders = build_hf_headers() # handles authentication\njsonld = requests.get(\"https://huggingface.co/api/datasets/open-r1/OpenR1-Math-220k/croissant\", headers=headers).json()\nds = Dataset(jsonld=jsonld)\nrecords = ds.records(\"all\")"},{"config_name":"extended","arguments":{"record_set":"extended","partial":false},"code":"import requests\nfrom huggingface_hub.file_download import build_hf_headers\nfrom mlcroissant import Dataset\n\n# Login using e.g. `huggingface-cli login` to access this dataset\nheaders = build_hf_headers() # handles authentication\njsonld = requests.get(\"https://huggingface.co/api/datasets/open-r1/OpenR1-Math-220k/croissant\", headers=headers).json()\nds = Dataset(jsonld=jsonld)\nrecords = ds.records(\"extended\")"}]},{"language":"python","library":"polars","function":"pl.read_parquet","loading_codes":[{"config_name":"all","arguments":{"splits":{"train":"all/train-*.parquet"}},"code":"import polars as pl\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = pl.read_parquet('hf://datasets/open-r1/OpenR1-Math-220k/all/train-*.parquet')\n"},{"config_name":"default","arguments":{"splits":{"train":"data/train-*.parquet"}},"code":"import polars as pl\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = pl.read_parquet('hf://datasets/open-r1/OpenR1-Math-220k/data/train-*.parquet')\n"},{"config_name":"extended","arguments":{"splits":{"train":"extended/train-*.parquet"}},"code":"import polars as pl\n\n# Login using e.g. `huggingface-cli login` to access this dataset\ndf = pl.read_parquet('hf://datasets/open-r1/OpenR1-Math-220k/extended/train-*.parquet')\n"}]}]}"><div class="relative md:w-full xl:w-auto xl:flex-none"> <button class="!from-gray-800 max-xl:mb-2 !to-black !text-white !gap-1 !border-gray-800 dark:!border-gray-900 btn w-full cursor-pointer text-sm" type="button"> <svg class="mr-1.5 !mr-0.5 " xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32" style="transform: rotate(360deg);"><path d="M31 16l-7 7l-1.41-1.41L28.17 16l-5.58-5.59L24 9l7 7z" fill="currentColor"></path><path d="M1 16l7-7l1.41 1.41L3.83 16l5.58 5.59L8 23l-7-7z" fill="currentColor"></path><path d="M12.419 25.484L17.639 6l1.932.518L14.35 26z" fill="currentColor"></path></svg> Use this dataset <svg class="-mr-1 text-gray-500" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 24 24"><path d="M16.293 9.293L12 13.586L7.707 9.293l-1.414 1.414L12 16.414l5.707-5.707z" fill="currentColor"></path></svg></button> </div> </div> </div> <div class="divider-column-vertical"></div> <div class="flex flex-col flex-wrap xl:flex-row"><a class="group mb-1.5 flex max-w-full flex-col overflow-hidden rounded-lg border border-gray-100 bg-gradient-to-r from-white via-white to-white px-2 py-1 hover:from-gray-50 dark:from-gray-900 dark:to-gray-925 dark:hover:to-gray-900 md:mr-1.5 pointer-events-none" rel="nofollow" target="_blank"><div class="truncate text-xs text-gray-400">Size of downloaded dataset files:</div> <div class="truncate text-sm "> 8.44 GB</div></a><a class="group mb-1.5 flex max-w-full flex-col overflow-hidden rounded-lg border border-gray-100 bg-gradient-to-r from-white via-white to-white px-2 py-1 hover:from-gray-50 dark:from-gray-900 dark:to-gray-925 dark:hover:to-gray-900 md:mr-1.5 " href="/datasets/open-r1/OpenR1-Math-220k/tree/refs%2Fconvert%2Fparquet/" rel="nofollow" target="_blank"><div class="truncate text-xs text-gray-400">Size of the auto-converted Parquet files:</div> <div class="truncate text-sm group-hover:underline"> 8.44 GB</div></a><a class="group mb-1.5 flex max-w-full flex-col overflow-hidden rounded-lg border border-gray-100 bg-gradient-to-r from-white via-white to-white px-2 py-1 hover:from-gray-50 dark:from-gray-900 dark:to-gray-925 dark:hover:to-gray-900 md:mr-1.5 pointer-events-none" rel="nofollow" target="_blank"><div class="truncate text-xs text-gray-400">Number of rows:</div> <div class="truncate text-sm "> 450,258</div></a></div> <div class="divider-column-vertical"></div> <h2 class="mb-5 flex items-baseline overflow-hidden whitespace-nowrap text-smd font-semibold text-gray-800"><svg class="mr-1.5 text-sm inline self-center flex-none text-gray-400" style="" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 24 24"><path class="uim-quaternary" d="M20.23 7.24L12 12L3.77 7.24a1.98 1.98 0 0 1 .7-.71L11 2.76c.62-.35 1.38-.35 2 0l6.53 3.77c.29.173.531.418.7.71z" opacity=".25" fill="currentColor"></path><path class="uim-tertiary" d="M12 12v9.5a2.09 2.09 0 0 1-.91-.21L4.5 17.48a2.003 2.003 0 0 1-1-1.73v-7.5a2.06 2.06 0 0 1 .27-1.01L12 12z" opacity=".5" fill="currentColor"></path><path class="uim-primary" d="M20.5 8.25v7.5a2.003 2.003 0 0 1-1 1.73l-6.62 3.82c-.275.13-.576.198-.88.2V12l8.23-4.76c.175.308.268.656.27 1.01z" fill="currentColor"></path></svg> Models trained or fine-tuned on <span class="ml-1 truncate font-mono text-[0.87rem] font-medium">open-r1/OpenR1-Math-220k</span></h2> <div class="space-y-3"><div class=""><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/open-r1/OpenR1-Qwen-7B"><div class="w-full truncate"><header class="flex items-center mb-1" title="open-r1/OpenR1-Qwen-7B"><img alt="" class="w-3 h-3 rounded-sm mr-1.5 flex-none flex-none" src="https://cdn-avatars.huggingface.co/v1/production/uploads/651e96991b97c9f33d26bde6/QG_uzb2VV3OeKO0c61ynh.png" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">open-r1/OpenR1-Qwen-7B</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"><svg class="mr-1.5 text-[.8rem] flex-none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" fill="currentColor" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 18 18"><path d="M16.2607 8.08202L14.468 6.28928C14.3063 6.12804 14.0873 6.03749 13.859 6.03749C13.6307 6.03749 13.4117 6.12804 13.25 6.28928L5.6375 13.904V16.9125H8.64607L16.2607 9.30002C16.422 9.13836 16.5125 8.91935 16.5125 8.69102C16.5125 8.4627 16.422 8.24369 16.2607 8.08202V8.08202ZM8.1953 15.825H6.725V14.3547L11.858 9.22118L13.3288 10.6915L8.1953 15.825ZM14.0982 9.92262L12.6279 8.45232L13.8606 7.21964L15.3309 8.68994L14.0982 9.92262Z"></path><path d="M6.18125 9.84373H7.26875V6.03748H8.9V4.94998H4.55V6.03748H6.18125V9.84373Z"></path><path d="M4.55 11.475H2.375V2.775H11.075V4.95H12.1625V2.775C12.1625 2.48658 12.0479 2.20997 11.844 2.00602C11.64 1.80208 11.3634 1.6875 11.075 1.6875H2.375C2.08658 1.6875 1.80997 1.80208 1.60602 2.00602C1.40207 2.20997 1.2875 2.48658 1.2875 2.775V11.475C1.2875 11.7634 1.40207 12.04 1.60602 12.244C1.80997 12.4479 2.08658 12.5625 2.375 12.5625H4.55V11.475Z"></path></svg> Text Generation <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <span class="truncate">Updated <time datetime="2025-02-11T08:10:35" title="Tue, 11 Feb 2025 08:10:35 GMT">9 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 5.04k <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32" fill="currentColor"><path d="M22.45,6a5.47,5.47,0,0,1,3.91,1.64,5.7,5.7,0,0,1,0,8L16,26.13,5.64,15.64a5.7,5.7,0,0,1,0-8,5.48,5.48,0,0,1,7.82,0L16,10.24l2.53-2.58A5.44,5.44,0,0,1,22.45,6m0-2a7.47,7.47,0,0,0-5.34,2.24L16,7.36,14.89,6.24a7.49,7.49,0,0,0-10.68,0,7.72,7.72,0,0,0,0,10.82L16,29,27.79,17.06a7.72,7.72,0,0,0,0-10.82A7.49,7.49,0,0,0,22.45,4Z"></path></svg> 30 </div></div> </a></article> </div><div class=""><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/mradermacher/OpenR1-Qwen-7B-GGUF"><div class="w-full truncate"><header class="flex items-center mb-1" title="mradermacher/OpenR1-Qwen-7B-GGUF"><img alt="" class="w-3 h-3 rounded-full mr-1.5 flex-none flex-none" src="/avatars/6b97d30ff0bdb5d5c633ba850af739cd.svg" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">mradermacher/OpenR1-Qwen-7B-GGUF</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"> <span class="truncate">Updated <time datetime="2025-02-11T18:48:13" title="Tue, 11 Feb 2025 18:48:13 GMT">9 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 2.21k <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32" fill="currentColor"><path d="M22.45,6a5.47,5.47,0,0,1,3.91,1.64,5.7,5.7,0,0,1,0,8L16,26.13,5.64,15.64a5.7,5.7,0,0,1,0-8,5.48,5.48,0,0,1,7.82,0L16,10.24l2.53-2.58A5.44,5.44,0,0,1,22.45,6m0-2a7.47,7.47,0,0,0-5.34,2.24L16,7.36,14.89,6.24a7.49,7.49,0,0,0-10.68,0,7.72,7.72,0,0,0,0,10.82L16,29,27.79,17.06a7.72,7.72,0,0,0,0-10.82A7.49,7.49,0,0,0,22.45,4Z"></path></svg> 1 </div></div> </a></article> </div><div class=""><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/brittlewis12/Mistral-Small-24B-Instruct-2501-reasoning-GGUF"><div class="w-full truncate"><header class="flex items-center mb-1" title="brittlewis12/Mistral-Small-24B-Instruct-2501-reasoning-GGUF"><img alt="" class="w-3 h-3 rounded-full mr-1.5 flex-none flex-none" src="/avatars/6b1d61e59837760e5d3b026ffee5117f.svg" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">brittlewis12/Mistral-Small-24B-Instruct-2501-reasoning-GGUF</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"><svg class="mr-1.5 text-[.8rem] flex-none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" fill="currentColor" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 18 18"><path d="M16.2607 8.08202L14.468 6.28928C14.3063 6.12804 14.0873 6.03749 13.859 6.03749C13.6307 6.03749 13.4117 6.12804 13.25 6.28928L5.6375 13.904V16.9125H8.64607L16.2607 9.30002C16.422 9.13836 16.5125 8.91935 16.5125 8.69102C16.5125 8.4627 16.422 8.24369 16.2607 8.08202V8.08202ZM8.1953 15.825H6.725V14.3547L11.858 9.22118L13.3288 10.6915L8.1953 15.825ZM14.0982 9.92262L12.6279 8.45232L13.8606 7.21964L15.3309 8.68994L14.0982 9.92262Z"></path><path d="M6.18125 9.84373H7.26875V6.03748H8.9V4.94998H4.55V6.03748H6.18125V9.84373Z"></path><path d="M4.55 11.475H2.375V2.775H11.075V4.95H12.1625V2.775C12.1625 2.48658 12.0479 2.20997 11.844 2.00602C11.64 1.80208 11.3634 1.6875 11.075 1.6875H2.375C2.08658 1.6875 1.80997 1.80208 1.60602 2.00602C1.40207 2.20997 1.2875 2.48658 1.2875 2.775V11.475C1.2875 11.7634 1.40207 12.04 1.60602 12.244C1.80997 12.4479 2.08658 12.5625 2.375 12.5625H4.55V11.475Z"></path></svg> Text Generation <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <span class="truncate">Updated <time datetime="2025-02-17T21:28:17" title="Mon, 17 Feb 2025 21:28:17 GMT">2 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 1.06k <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32" fill="currentColor"><path d="M22.45,6a5.47,5.47,0,0,1,3.91,1.64,5.7,5.7,0,0,1,0,8L16,26.13,5.64,15.64a5.7,5.7,0,0,1,0-8,5.48,5.48,0,0,1,7.82,0L16,10.24l2.53-2.58A5.44,5.44,0,0,1,22.45,6m0-2a7.47,7.47,0,0,0-5.34,2.24L16,7.36,14.89,6.24a7.49,7.49,0,0,0-10.68,0,7.72,7.72,0,0,0,0,10.82L16,29,27.79,17.06a7.72,7.72,0,0,0,0-10.82A7.49,7.49,0,0,0,22.45,4Z"></path></svg> 2 </div></div> </a></article> </div><div class=""><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/mradermacher/Open-R1-Math-7B-Instruct-i1-GGUF"><div class="w-full truncate"><header class="flex items-center mb-1" title="mradermacher/Open-R1-Math-7B-Instruct-i1-GGUF"><img alt="" class="w-3 h-3 rounded-full mr-1.5 flex-none flex-none" src="/avatars/6b97d30ff0bdb5d5c633ba850af739cd.svg" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">mradermacher/Open-R1-Math-7B-Instruct-i1-GGUF</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"> <span class="truncate">Updated <time datetime="2025-02-12T12:32:56" title="Wed, 12 Feb 2025 12:32:56 GMT">8 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 993 </div></div> </a></article> </div><div class="hidden md:block"><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/mradermacher/OpenR1-Qwen-7B-i1-GGUF"><div class="w-full truncate"><header class="flex items-center mb-1" title="mradermacher/OpenR1-Qwen-7B-i1-GGUF"><img alt="" class="w-3 h-3 rounded-full mr-1.5 flex-none flex-none" src="/avatars/6b97d30ff0bdb5d5c633ba850af739cd.svg" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">mradermacher/OpenR1-Qwen-7B-i1-GGUF</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"> <span class="truncate">Updated <time datetime="2025-02-11T19:22:24" title="Tue, 11 Feb 2025 19:22:24 GMT">9 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 910 </div></div> </a></article> </div><div class="hidden md:block"><article class="overview-card-wrapper group/repo "><a class="flex items-center justify-between gap-4 p-2" href="/yentinglin/Mistral-Small-24B-Instruct-2501-reasoning"><div class="w-full truncate"><header class="flex items-center mb-1" title="yentinglin/Mistral-Small-24B-Instruct-2501-reasoning"><img alt="" class="w-3 h-3 rounded-full mr-1.5 flex-none flex-none" src="https://cdn-avatars.huggingface.co/v1/production/uploads/5df9c78eda6d0311fd3d541f/8oDFuP77l5zhvamXNVmnc.jpeg" crossorigin="anonymous"> <h4 class="text-md truncate font-mono text-black dark:group-hover/repo:text-yellow-500 group-hover/repo:text-indigo-600 text-sm">yentinglin/Mistral-Small-24B-Instruct-2501-reasoning</h4> </header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"><svg class="mr-1.5 text-[.8rem] flex-none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" fill="currentColor" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 18 18"><path d="M16.2607 8.08202L14.468 6.28928C14.3063 6.12804 14.0873 6.03749 13.859 6.03749C13.6307 6.03749 13.4117 6.12804 13.25 6.28928L5.6375 13.904V16.9125H8.64607L16.2607 9.30002C16.422 9.13836 16.5125 8.91935 16.5125 8.69102C16.5125 8.4627 16.422 8.24369 16.2607 8.08202V8.08202ZM8.1953 15.825H6.725V14.3547L11.858 9.22118L13.3288 10.6915L8.1953 15.825ZM14.0982 9.92262L12.6279 8.45232L13.8606 7.21964L15.3309 8.68994L14.0982 9.92262Z"></path><path d="M6.18125 9.84373H7.26875V6.03748H8.9V4.94998H4.55V6.03748H6.18125V9.84373Z"></path><path d="M4.55 11.475H2.375V2.775H11.075V4.95H12.1625V2.775C12.1625 2.48658 12.0479 2.20997 11.844 2.00602C11.64 1.80208 11.3634 1.6875 11.075 1.6875H2.375C2.08658 1.6875 1.80997 1.80208 1.60602 2.00602C1.40207 2.20997 1.2875 2.48658 1.2875 2.775V11.475C1.2875 11.7634 1.40207 12.04 1.60602 12.244C1.80997 12.4479 2.08658 12.5625 2.375 12.5625H4.55V11.475Z"></path></svg> Text Generation <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <span class="truncate">Updated <time datetime="2025-02-17T06:42:29" title="Mon, 17 Feb 2025 06:42:29 GMT">3 days ago</time></span> <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-0.5" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" viewBox="0 0 32 32"><path fill="currentColor" d="M26 24v4H6v-4H4v4a2 2 0 0 0 2 2h20a2 2 0 0 0 2-2v-4zm0-10l-1.41-1.41L17 20.17V2h-2v18.17l-7.59-7.58L6 14l10 10l10-10z"></path></svg> 811 <span class="px-1.5 text-gray-300 dark:text-gray-500">• </span> <svg class="flex-none w-3 text-gray-400 mr-1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" focusable="false" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 32 32" fill="currentColor"><path d="M22.45,6a5.47,5.47,0,0,1,3.91,1.64,5.7,5.7,0,0,1,0,8L16,26.13,5.64,15.64a5.7,5.7,0,0,1,0-8,5.48,5.48,0,0,1,7.82,0L16,10.24l2.53-2.58A5.44,5.44,0,0,1,22.45,6m0-2a7.47,7.47,0,0,0-5.34,2.24L16,7.36,14.89,6.24a7.49,7.49,0,0,0-10.68,0,7.72,7.72,0,0,0,0,10.82L16,29,27.79,17.06a7.72,7.72,0,0,0,0-10.82A7.49,7.49,0,0,0,22.45,4Z"></path></svg> 37 </div></div> </a></article> </div> <a class="flex items-center pt-1 text-sm text-gray-500 underline hover:text-gray-700" href="/models?dataset=dataset:open-r1/OpenR1-Math-220k">Browse 45 models trained on this dataset </a></div> <div class="divider-column-vertical"></div> <h2 class="mb-5 flex items-baseline overflow-hidden whitespace-nowrap text-smd font-semibold text-gray-800"><svg class="mr-1 inline self-center flex-none" width="1em" height="1em" aria-hidden="true" focusable="false" role="img" viewBox="0 0 12 13" fill="none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"><rect x="2" y="2.49902" width="8" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.4"></rect><rect x="6.21875" y="6.7334" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.7"></rect><rect x="2" y="6.73438" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.5"></rect></svg> Collections including <span class="ml-1 truncate font-mono text-[0.86rem] font-medium">open-r1/OpenR1-Math-220k</span></h2> <div class="flex flex-col gap-3"> <div class="flex flex-col"><article class="overview-card-wrapper group/collection relative !rounded-md !from-white !to-white dark:!from-gray-900 dark:!to-gray-900 peer"><a href="/collections/open-r1/reasoning-datasets-67980cac6e816a0eda98c678" class="block p-2"><header class="mb-0.5 flex items-center" title="🧠 Reasoning datasets"><h4 class="text-md truncate text-smd font-semibold group-hover/collection:underline">🧠 Reasoning datasets</h4> <div class="ml-2 flex items-center rounded sm:ml-2.5 bg-indigo-500/10 py-0.5 pl-1 pr-1.5 text-xs leading-none text-gray-700"><svg class="mr-0.5 flex-none text-indigo-700" width="1em" height="1em" aria-hidden="true" focusable="false" role="img" viewBox="0 0 12 13" fill="none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"><rect x="2" y="2.49902" width="8" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.4"></rect><rect x="6.21875" y="6.7334" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.7"></rect><rect x="2" y="6.73438" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.5"></rect></svg> Collection </div></header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"><span class="max-w-64 truncate">Datasets with reasoning traces for math and code released by the community</span> <span class="px-1.5 text-gray-300">• </span> <span>12 items</span> <span class="px-1.5 text-gray-300">• </span> <span class="truncate">Updated <time datetime="2025-02-20T09:05:50" title="2025-02-20T09:05:50.364Z">21 minutes ago</time></span> <span class="px-1.5 text-gray-300">•</span> <svg class="flex-none w-3 mr-1 text-gray-400" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 12 12" fill="transparent"><path d="M9.30013 9.29152H9.3H2.7H2.69987C2.62308 9.29154 2.54762 9.27146 2.481 9.23328C2.41437 9.1951 2.3589 9.14015 2.32009 9.07389C2.28128 9.00763 2.26048 8.93237 2.25977 8.85558C2.25907 8.7798 2.27796 8.70513 2.31458 8.63882L5.62238 2.9426L5.67518 2.85168C5.7059 2.81806 5.74178 2.78928 5.78164 2.76649C5.84813 2.72848 5.9234 2.70848 6 2.70848C6.0766 2.70848 6.15187 2.72848 6.21836 2.76649C6.28441 2.80425 6.33953 2.85848 6.37836 2.92389L9.68527 8.63855C9.72199 8.70493 9.74093 8.7797 9.74023 8.85558C9.73952 8.93237 9.71872 9.00763 9.67991 9.07389C9.6411 9.14015 9.58563 9.1951 9.519 9.23328C9.45238 9.27146 9.37692 9.29154 9.30013 9.29152Z" stroke="currentColor"></path></svg> 74</div></a></article> <div class="overview-card-wrapper -z-1 !rounded-md !bg-gradient-to-t -mt-[1.60rem] h-8 scale-x-[98%] !from-gray-50 !to-gray-50 !shadow-none transition-transform peer-hover:-translate-y-[2px] peer-hover:from-gray-100/80 dark:!from-gray-925 dark:!to-gray-925 dark:peer-hover:from-black"></div></div> <div class="flex flex-col"><article class="overview-card-wrapper group/collection relative !rounded-md !from-white !to-white dark:!from-gray-900 dark:!to-gray-900 peer"><a href="/collections/open-r1/openr1-math-67ab097f8087b6634035e764" class="block p-2"><header class="mb-0.5 flex items-center" title="OpenR1-Math"><h4 class="text-md truncate text-smd font-semibold group-hover/collection:underline">OpenR1-Math</h4> <div class="ml-2 flex items-center rounded sm:ml-2.5 bg-indigo-500/10 py-0.5 pl-1 pr-1.5 text-xs leading-none text-gray-700"><svg class="mr-0.5 flex-none text-indigo-700" width="1em" height="1em" aria-hidden="true" focusable="false" role="img" viewBox="0 0 12 13" fill="none" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"><rect x="2" y="2.49902" width="8" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.4"></rect><rect x="6.21875" y="6.7334" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.7"></rect><rect x="2" y="6.73438" width="3.78055" height="3.76425" rx="1.16774" fill="currentColor" fill-opacity="0.5"></rect></svg> Collection </div></header> <div class="mr-1 flex items-center overflow-hidden whitespace-nowrap text-sm leading-tight text-gray-400"><span class="max-w-64 truncate">Dataset and SFT model distilled from DeepSeek-R1. Check out our blog post for more details: https://huggingface.co/blog/open-r1/update-2</span> <span class="px-1.5 text-gray-300">• </span> <span>3 items</span> <span class="px-1.5 text-gray-300">• </span> <span class="truncate">Updated <time datetime="2025-02-14T18:14:35" title="2025-02-14T18:14:35.817Z">6 days ago</time></span> <span class="px-1.5 text-gray-300">•</span> <svg class="flex-none w-3 mr-1 text-gray-400" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-hidden="true" role="img" width="1em" height="1em" preserveAspectRatio="xMidYMid meet" viewBox="0 0 12 12" fill="transparent"><path d="M9.30013 9.29152H9.3H2.7H2.69987C2.62308 9.29154 2.54762 9.27146 2.481 9.23328C2.41437 9.1951 2.3589 9.14015 2.32009 9.07389C2.28128 9.00763 2.26048 8.93237 2.25977 8.85558C2.25907 8.7798 2.27796 8.70513 2.31458 8.63882L5.62238 2.9426L5.67518 2.85168C5.7059 2.81806 5.74178 2.78928 5.78164 2.76649C5.84813 2.72848 5.9234 2.70848 6 2.70848C6.0766 2.70848 6.15187 2.72848 6.21836 2.76649C6.28441 2.80425 6.33953 2.85848 6.37836 2.92389L9.68527 8.63855C9.72199 8.70493 9.74093 8.7797 9.74023 8.85558C9.73952 8.93237 9.71872 9.00763 9.67991 9.07389C9.6411 9.14015 9.58563 9.1951 9.519 9.23328C9.45238 9.27146 9.37692 9.29154 9.30013 9.29152Z" stroke="currentColor"></path></svg> 6</div></a></article> <div class="overview-card-wrapper -z-1 !rounded-md !bg-gradient-to-t -mt-[1.60rem] h-8 scale-x-[98%] !from-gray-50 !to-gray-50 !shadow-none transition-transform peer-hover:-translate-y-[2px] peer-hover:from-gray-100/80 dark:!from-gray-925 dark:!to-gray-925 dark:peer-hover:from-black"></div></div></div> <div class="divider-column-vertical md:hidden"></div></section></div></main> <footer class="b-12 mb-2 flex border-t border-gray-100 md:h-14"><nav class="container relative flex flex-col justify-between space-y-2 py-6 text-gray-500 *:max-md:self-start md:flex-row md:items-center md:space-y-0 md:py-0 md:text-sm"><div class="SVELTE_HYDRATER contents" 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