{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"mmlu-pro","formal_name":"MMLU-Pro","introduction":"MMLU-Pro rebuilds MMLU with ten answer options instead of four and removes items that no longer separate models. Questions span fourteen academic and professional subjects.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"897656d1-f4f6-587a-9bc6-27005d17e1da","task_key":"default--test--8580","task_revision_id":"1","upstream_id":"8580","short_description":"Water drips out of a hole at the vertex of an upside down cone at a rate of 3…","config":"default","split":"test","body":"{\"category\":\"math\",\"options\":\"['−28/(3π) cm/min', '−8/(3π) cm/min', '−24/(3π) cm/min', '−48/π cm/min', '−12/(3π) cm/min', '−4/(3π) cm/min', '−32/(3π) cm/min', '−20/(3π) cm/min', '−40/(3π) cm/min', '−16/(3π) cm/min']\",\"question\":\"Water drips out of a hole at the vertex of an upside down cone at a rate of 3 cm^3 per minute. The\\ncone’s height and radius are 2 cm and 1 cm, respectively. At what rate does the height of the water change\\nwhen the water level is half a centimeter below the top of the cone? The volume of a cone is V = (π/3)*r^2*h,\\nwhere r is the radius and h is the height of the cone.\",\"src\":\"ori_mmlu-college_mathematics\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}