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8 lines
No EOL
2.9 KiB
JSON
8 lines
No EOL
2.9 KiB
JSON
{
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"question": "Maria buys 8 shares of a stock for $8 each. The stock price increases 50% the first year Maria holds it, then decreases 25% in the second year. What is the final value of all Maria's shares?",
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"answer": "First find the initial total value of Maria's purchase: 8 shares * $8/share = $<<8*8=64>>64\nThen find the amount of the first price increase: $64 * .5 = $<<64*.5=32>>32\nAdd that amount to the initial value to find the value after the first year: $64 + $32 = $<<64+32=96>>96\nThen multiply that amount by 25% to find the amount of the decrease in the second year: $96 * 25% = $<<96*25*.01=24>>24\nThen subtract that amount from the value after the first year to find the final value: $96 - $24 = $<<96-24=72>>72\n#### 72",
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"id_orig": 336,
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"id_shuffled": 84,
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"question_annotated": "{name,Maria} buys {n,8} shares of a stock for {cur,$}{p,8} each. The stock price increases {r1,50}% in the first year {name,Maria} holds it, then decreases {r2,25}% in the second year. In the third year, the stock price increases by {r3,10}%. By the end of the third year, how much profit has {name,Maria} made?\n\n#init:\n- name = sample(names_female)\n- cur = sample(currencies_sym)\n- $n = range(5, 20)\n- $p = range(5, 100)\n- $r1 = range(10, 100, 5)\n- $r2 = range(5, 51, 5)\n- $r3 = range(5, 26, 5)\n\n\n#conditions:\n- is_int(n * p * r1/100)\n- is_int(n * p * (1 + r1/100) * (1 - r2/100))\n- is_int(n * p * (1 + r1/100) * (1 - r2/100) * (1+r3/100))\n- int(n * p * (1 + r1/100) * (1 - r2/100) * (1+r3/100)) - n * p > 10\n\n#answer: int(n * p * (1 + r1/100) * (1 - r2/100) * (1+r3/100)) - n * p",
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"answer_annotated": "First find the initial total value of {name}'s purchase: {n} shares * {cur}{p}/share = {cur}<<{n}*{p}={n*p}>>{n*p}\nThen find the amount of the first price increase: {cur}{n*p} * {r1/100} = {cur}<<{n*p}*{r1/100}={n*p*r1/100}>>{int(n*p*r1/100)}\nAdd that amount to the initial value to find the value after the first year: {cur}{n*p} + {cur}{n*p*r1/100} = {cur}<<{n*p}+{n*p*r1/100}={n*p*(1+r1/100)}>>{int(n*p*(1+r1/100))}\nThen multiply that amount by {r2}% to find the amount of the decrease in the second year: {cur}{n*p*(1+r1/100)} * {r2}% = {cur}<<{n*p*(1+r1/100)}*{r2}*.01={n*p*(1+r1/100)*r2/100}>>{int(n*p*(1+r1/100)*r2/100)}\nThen subtract that amount from the value after the first year to find the final value: {cur}{int(n*p*(1+r1/100))} - {cur}{int(n*p*(1+r1/100)*r2/100)} = {cur}<<{int(n*p*(1+r1/100))}-{int(n*p*(1+r1/100)*r2/100)}={int(n*p*(1+r1/100)*(1-r2/100))}>>{int(n*p*(1+r1/100)*(1-r2/100))}. \nIn the third year the stocks increase {r3}%, so we have {int(n*p*(1+r1/100)*(1-r2/100))} * {100+r3}% = {int(n*p*(1+r1/100)*(1-r2/100)*(1+r3/100))}. \nThe total profit is {cur}{int(n*p*(1+r1/100)*(1-r2/100)*(1+r3/100))} - {cur}{n * p} = {cur}{ int(n * p * (1 + r1/100) * (1 - r2/100) * (1+r3/100)) - n * p} \n#### {int(n * p * (1 + r1/100) * (1 - r2/100) * (1+r3/100)) - n * p}"
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} |