Problem 7 · 2019 AMC 8
Easy
Arithmetic & Operations
sum-constraintestimate-and-pick
Shauna takes five tests, each worth a maximum of 100 points. Her scores on the first three tests are 76, 94, and 87. In order to average 81 for all five tests, what is the lowest score she could earn on one of the other two tests?
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Answer: A — 48.
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Hint 1 of 2
The two unknown scores must add to a fixed total. If a fixed amount is split between two scores, making one as small as possible means making the other as large as possible — push the partner to its ceiling of 100.
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Hint 2 of 2
Average 81 over 5 tests means the scores total 5 × 81 = 405. Subtract the three known scores to see what the last two must add to, then give one of them the maximum 100.
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Approach: fixed sum — shove one score to its max so the other is smallest
- An average of 81 across 5 tests means a total of 5 × 81 = 405. The first three give 76 + 94 + 87 = 257, so the last two must sum to 405 − 257 = 148.
- Those two scores share a fixed 148. To make one as low as possible, make its partner as high as possible: 100 (the test maximum).
- Lowest possible = 148 − 100 = 48.
- Why this transfers: "minimize one of several things whose total is fixed" is always solved by maxing out everything else — here the cap of 100 is what makes 48 reachable rather than going lower.
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