2012 AMC 8 Problem 7

Attempt Problem 7 of the 2012 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

Isabella must take four 100100-point tests in her math class. Her goal is to achieve an average grade of at least 9595 on the tests. Her first two test scores were 9797 and 91.91. After seeing her score on the third test, she realized that she could still reach her goal. What is the lowest possible score she could have made on the third test?
90 90
92 92
95 95
96 96
97 97
Answer: B
Concepts:meanoptimization
Small hint:
A 9595 average on four tests requires a total of 4⋅954\cdot95.
Big hint:
Make the fourth test as large as possible to minimize the third score.
Video solution:
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Written solution:
If the average is 95,95, then the sum of the tests is 95⋅4=380.95\cdot 4 = 380 . Since we have the first two tests, the sum of the last two tests is 380380 minus the first two scores.
This makes the sum of the last two scores equal to 380−91−97=192.380-91-97 = 192. Her last two scores therefore have a sum of 192.192.
Given the sum of the tests we try to minimize one score, then we must maximize the other test. Therefore, we maximize the fourth test by making it 100.100. This would make the third test equal to 192−100=92.192 -100 = 92.
Thus, the answer is B .
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