Problem 14 · 2017 AMC 8
Medium
Fractions, Decimals & Percents
percent-multipliersubstitution
Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only 80% of the problems she solved alone, but overall 88% of her answers were correct. Zoe had correct answers to 90% of the problems she solved alone. What was Zoe's overall percentage of correct answers?
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Answer: C — 93%.
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Hint 1 of 2
The hidden link is the 'together' half — the girls worked it side by side, so they got the same questions right there. Chloe's data secretly tells you that shared score, which is the missing piece for Zoe.
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Hint 2 of 2
Pick a friendly total of 100 problems (50 alone + 50 together). Percentages with no given count are free to scale, so choose the number that makes the arithmetic clean.
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Approach: pick 100 problems, then mine Chloe's data for the shared half
- Let there be 100 problems: 50 alone + 50 together. (No count was given, so we're free to choose the convenient one — the percentages won't change.)
- Chloe alone: 80% of 50 = 40 correct. Chloe overall: 88 of 100. So her together-score = 88 − 40 = 48 of 50. And that 48 is shared — it's also Zoe's together-score.
- Zoe alone: 90% of 50 = 45 correct. Zoe total = 45 + 48 = 93 of 100 = 93%.
- Why this transfers: when two people share part of a task, that shared part is one number you can solve for from either person and then reuse for the other.
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