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8 lines
No EOL
4 KiB
JSON
8 lines
No EOL
4 KiB
JSON
{
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"question": "Artie has a flower stand at the Farmers Market. He sells three kinds of flowers: marigolds, petunias and begonias. He usually sells marigolds for $2.74 per pot, petunias for $1.87 per pot and begonias for $2.12 per pot. Artie has no change today, so he has decided to round all his prices to the nearest dollar. If Artie sells 12 pots of marigolds, 9 pots of petunias and 17 pots of begonias, how much will he make?",
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"answer": "Artie will round his marigolds up from $2.74 to $3, since the number following the 2 is 5 or higher.\nArtie will round his petunias up from $1.87 to $2, since the number following the 1 is 5 or higher.\nArtie will round his begonias down from $2.12 to $2, since the number following the 2 is less than 5.\nArtie sells 12 marigolds x $3 = $<<12*3=36>>36\nArtie sells 9 petunias x $2 = $<<9*2=18>>18\nArtie sells 17 begonias x $2 = $<<17*2=34>>34\nAltogether, Artie will make $36 + $18 + $34 = $<<36+18+34=88>>88\n#### 88",
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"id_orig": 74,
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"id_shuffled": 51,
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"question_annotated": "{name,Artie} has a {product,flower} stand at the {location,Farmers Market}. {name,Artie} sells three kinds of {product,flower}s: {item1,marigolds}, {item2,petunias} and {item3,begonias}. {name,Artie} usually sells {item1,marigolds} for {cur,$}{p1,2.74} per {unit,pot}, {item2,petunias} for {cur,$}{p2,1.87} per {unit,pot} and {item3,begonias} for {cur,$}{p3,2.12} per {unit,pot}. {name,Artie} has no change today, so {name,Artie} has decided to round all his prices to the nearest dollar. {name,Artie} sells {n1,12} {unit,pot}s of {item1,marigolds}, {n2,9} {unit,pot}s of {item2,petunias} and {n3,17} {unit,pot}s of {item3,begonias}. How much will {name,Artie} make, after receiving a {k, 10}% tip on every purchase?\n\n#init:\n- name = sample(names)\n- product = sample([\"flower\", \"herb\", \"plant\"])\n- location = sample([\"local fair\", \"community market\", \"street bazaar\", \"town square\"])\n- item1, item2, item3 = sample([\"roses\", \"daisies\", \"tulips\", \"lilies\", \"sunflowers\", \"orchids\"], 3)\n- unit = sample([\"bunch\", \"basket\", \"bouquet\", \"bundle\"])\n- cur = sample(currencies_sym)\n- $k = range(5, 26, 5)\n- $p1 = fix_floats(np.arange(1.26, 3.53, 0.33))\n- $p2 = fix_floats(np.arange(2.27, 5.53, 0.26))\n- $p3 = fix_floats(np.arange(4.85, 6.53, 0.31))\n- $n1 = range(5, 21)\n- $n2 = range(15, 31)\n- $n3 = range(35, 41)\n\n#conditions:\n- is_int((n1*round(p1) + n2*round(p2) + n3*round(p3))*(k/100.0))\n\n#answer: int((n1*round(p1) + n2*round(p2) + n3*round(p3))*(1+(k/100.0)))",
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"answer_annotated": "{name} will round his {item1} up from {cur}{p1} to {cur}{round(p1)}, since the number following the {int(p1)} is 5 or higher.\n{name} will round his {item2} {'up' if round(p2) > p2 else 'down'} from {cur}{p2} to {cur}{round(p2)}, since the number following the {int(p2)} is {'5 or higher' if round(p2) > p2 else 'less than 5'}.\n{name} will round his {item3} {'up' if round(p3) > p3 else 'down'} from {cur}{p3} to {cur}{round(p3)}, since the number following the {int(p3)} is {'5 or higher' if round(p3) > p3 else 'less than 5'}.\n{name} sells {n1} {item1} x {cur}{round(p1)} = {cur}<<{n1}*{round(p1)}={n1*round(p1)}>>{n1*round(p1)}\n{name} sells {n2} {item2} x {cur}{round(p2)} = {cur}<<{n2}*{round(p2)}={n2*round(p2)}>>{n2*round(p2)}\n{name} sells {n3} {item3} x {cur}{round(p3)} = {cur}<<{n3}*{round(p3)}={n3*round(p3)}>>{n3*round(p3)}\nAltogether, {name} will make {cur}{n1*round(p1)} + {cur}{n2*round(p2)} + {cur}{n3*round(p3)} = {cur}<<{n1*round(p1)}+{n2*round(p2)}+{n3*round(p3)}={n1*round(p1) + n2*round(p2) + n3*round(p3)}>>{n1*round(p1) + n2*round(p2) + n3*round(p3)}.\nFinally {name} made {k}% tip so we should add {k}% * {cur}{n1*round(p1) + n2*round(p2) + n3*round(p3)} = {cur}{int((n1*round(p1) + n2*round(p2) + n3*round(p3))*((k/100)))}. \nThe total will be {cur}{n1*round(p1) + n2*round(p2) + n3*round(p3)} + {cur}{int((n1*round(p1) + n2*round(p2) + n3*round(p3))*((k/100)))} = {cur}{int((n1*round(p1) + n2*round(p2) + n3*round(p3))*(1+(k/100.0)))}.\n#### {int((n1*round(p1) + n2*round(p2) + n3*round(p3))*(1+(k/100.0)))}"
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} |