open-thought/reasoning-gym
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abdulhakeem 2025-01-30 23:44:12 -06:00
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{
"question": "Cole hid 3 dozen eggs in the yard for the Easter egg hunt. Lamar finds 5 eggs. Stacy finds twice as many as Lamar. Charlie finds 2 less than Stacy. And Mei finds half as many as Charlie. How many eggs are still hidden in the yard?",
"answer": "Cole hides 3 x 12 = <<3*12=36>>36 eggs.\nLamar finds 5 eggs.\nStacy finds 5 x 2 = <<5*2=10>>10 eggs.\nCharlie finds 10 - 2 = <<10-2=8>>8 eggs.\nMei finds 8 / 2 = <<8/2=4>>4 eggs.\nThe children find a total of 5 + 10 + 8 + 4 = <<5+10+8+4=27>>27 eggs.\nThe total number of hidden eggs still in the yard is 36 - 27 = <<36-27=9>>9 eggs.\n#### 9",
"id_orig": 1164,
"id_shuffled": 47,
"question_annotated": "{name1,Cole} initially had {n,3} dozen {food,eggs}, then ate {eat,three} {food,eggs} for breakfast, and then hid the rest in the yard for the Easter egg hunt. {name2,Lamar} finds {x,5} {food,eggs}. {name3,Stacy} finds {mult1,twice} as many as {name2,Lamar}. {name4,Charlie} finds {y,2} less than {name3,Stacy}. {name5,Mei} finds {frac,half} as many as {name4,Charlie}, and {name6,Joe} finds {extra,two} dozen {food,eggs} more than {name3,Starcy} and {name2,Lamar} combined. How many {food,eggs} are still hidden in the yard?\n\n#init:\n- name1, name2, name3, name4, name5 , name6 = sample(names, 6)\n- food = sample(['eggs', 'waffles', 'pancakes', 'cookies'])\n- $n = range(2, 12)\n- $x = range(3, 15)\n- $y = range(1, 5)\n- $mult1 = sample(multiple_ice+multi_times)\n- $frac = sample(fractions)\n- $extra = numbers_within(2, 4)\n- $eat = range(2, 6)\n\n#conditions:\n- is_int((n * 12-eat) - (x + mult1 * x + (mult1 * x - y) + frac * (mult1 * x - y)))\n- is_int(x * mult1)\n- x * mult1 - y > 0\n- is_int((x * mult1 - y) * frac)\n- n * 12 - eat > x + mult1 * x + (mult1 * x - y) + frac * (mult1 * x - y) + ((12*extra)+x+mult1 * x)\n\n#answer: (n * 12-eat)- (x + mult1 * x + (mult1 * x - y) + frac * (mult1 * x - y)+((12*extra)+x+mult1 * x))",
"answer_annotated": "{name1} has {n} x 12 = <<{n}*12={n*12}>>{n*12} {food}. But {name1} ate {eat} of them. So {n*12}-{eat}=<<{n*12}-{eat}={n*12-eat}>>{n*12 - eat} {food} will be hidden for the Easter egg hunt.\n{name2} finds {x} {food}.\n{name3} finds {x} x {mult1} = <<{x}*{mult1}={x*mult1}>>{x*mult1} {food}.\n{name4} finds {x*mult1} - {y} = <<{x*mult1}-{y}={x*mult1-y}>>{x*mult1-y} {food}.\n{name5} finds {x*mult1-y} * {frac} = <<{x*mult1-y}*{frac}={(x*mult1-y)*frac}>>{(x*mult1-y)*frac} {food}.\n{name6} finds 12*{extra} + {x} + {mult1*x} = {(12*extra) + x + mult1*x}.\nThe children find a total of {x} + {x*mult1} + {x*mult1-y} + {(x*mult1-y)*frac} + {(12*extra) + x + mult1*x} = <<{x}+{x*mult1}+{x*mult1-y}+{(x*mult1-y)*frac}+{(12*extra) + x + mult1*x}={x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac+((12*extra) + x + mult1*x)}>>{x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac+((12*extra) + x + mult1*x)} {food}.\nThe total number of hidden {food} still in the yard is {n*12-eat} - {x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac + ((12*extra) + x + mult1*x)} = <<{n*12-eat}-{x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac + ((12*extra) + x + mult1*x)}={(n * 12 - eat) - (x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac + ((12*extra) +x+mult1 * x))}>>{(n * 12 - eat) - (x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac + ((12*extra) +x+mult1 * x))} {food}.\n#### {(n * 12 - eat) - (x + x*mult1 + (x*mult1-y) + (x*mult1-y)*frac + ((12*extra) +x+mult1 * x))}"
}