Still experimenting

reasoning_gym/data/gsm_data/p1/0058.json

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+{
+  "question": "One dwarf can mine 12 pounds of ore per day with his bare hands. He can mine twice as much with an iron pickaxe and 50% more with a steel pickaxe than with an iron pickaxe. How many pounds of ore can 40 dwarves with steel pickaxes mine in a month with 30 days?",
+  "answer": "First find how much ore a dwarf can mine with an iron pickaxe: 12 pounds/day * 2 = <<12*2=24>>24 pounds/day\nThen multiply that amount by 150% to find how much a dwarf can mine with a steel pickaxe: 24 pounds/day * 150% = <<24*150*.01=36>>36 pounds/day\nThen multiply the amount one dwarf can mine in a day with a steel pickaxe by the number of dwarves: 36 pounds/day/dwarf * 40 dwarves = <<36*40=1440>>1440 pounds/day\nThen multiply the daily amount of ore by the number of days to find the total ore mined in a month: 1440 pounds/day * 30 days = <<1440*30=43200>>43200 pounds/day\n#### 43200",
+  "id_orig": 1026,
+  "id_shuffled": 58,
+  "question_annotated": "One {worker,dwarf} can mine {base,12} {unit,pounds} of ore per day with {tool1,his bare hands}. He can mine {mult1,twice} as much with one {tool2,iron pickaxe} and {mult2,50}% more with a {tool3,steel pickaxe} than with one {tool2,iron pickaxe}. How many {unit,pounds} of ore can {n,40} {worker,dwarf}s mine in a month with {days,30} days if they all use {tool2,iron pickaxe} in the first half, and then switch to {tool3,steel pickaxe}s in the second half of the month?\n\n#init:\n- worker = sample([\"miner\", \"goblin\", \"gnome\", \"troll\"])\n- tool1 = sample([\"bare hands\", \"basic shovel\", \"wooden pickaxe\"])\n- unit = sample([\"pounds\", \"kgs\"])\n- tool2 = sample([\"iron pickaxe\", \"bronze pickaxe\", \"silver pickaxe\"])\n- tool3 = sample([\"steel pickaxe\", \"diamond pickaxe\", \"mithril pickaxe\"])\n- $base = range(5, 26)\n- $mult1 = sample(multi_times+multiple_ice)\n- $mult2 = range(30, 81, 5)\n- $n = range(8, 21)\n- $days = range(28, 37, 2)\n\n#conditions:\n- is_int(mult2/100*base*mult1)\n- int(base * mult1 * (1 + mult2/100) * n * days) < 6000\n\n#answer: int(base * mult1 * (1 + mult2/100) * n * days//2) + (n*days//2*base*mult1)",
+  "answer_annotated": "First find how much ore a {worker} can mine with an {tool2}: {base} pounds/day * {mult1} = <<{base}*{mult1}={base*mult1}>>{base*mult1} {unit}/day.\nThen multiply that amount by {100+mult2}% to find how much a {worker} can mine with a {tool3}: {base*mult1} {unit}/day * {100+mult2}% = <<{base*mult1}*{100+mult2}*.01={int(base*mult1*(1+mult2/100))}>>{int(base*mult1*(1+mult2/100))} {unit}/day.\nSince {worker}s use {tool2} for the first half of the month, we should count the amount of ores they mine using {tool2} in {days//2} days. {base*mult1} * {days//2} = {base*mult1*days//2} {unit}. They also use {tool3} in the next {days//2} days: {int(base*mult1*(1+mult2/100))}*{days//2}={int(base*mult1*(1+mult2/100)) * days//2}. \nIn total, each {worker} mines {base*mult1*days//2} {unit} with {tool2}, and {int(base*mult1*(1+mult2/100))*days//2} {unit} with {tool3} per day. In total, that would be {base*mult1*days//2}+{int(base*mult1*(1+mult2/100))*days//2}={(base*mult1*days//2) + (int(base*mult1*(1+mult2/100))*days//2)} {unit} per day. \nThen multiply the amount one {worker} can mine in a day by the number of {worker}s, which is {n}: {(base*mult1*days//2) + (int(base*mult1*(1+mult2/100))*days//2)}*{n}={n*((base*mult1*days//2) + (int(base*mult1*(1+mult2/100))*days//2))} {unit}.\n#### {int(base * mult1 * (1 + mult2/100) * n * days//2) + (n*days//2*base*mult1)}"
+}