2011 AMC 10B Problem 2

Attempt Problem 2 of the 2011 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

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

2.

Josanna’s test scores to date are 90,90, 80,80, 70,70, 60,60, and 85.85. Her goal is to raise her test average at least 33 points with her next test. What is the minimum test score she would need to accomplish this goal?
8080
8282
8585
9090
9595
Answer: E
Concepts:meanwork backwards
Small hint:
Find the current total score and target average
Big hint:
The sixth score must bring the total to 6⋅806\cdot80
Solution:
Her current average is 90+80+70+60+855=77,\dfrac{90+80+70+60+85}5 = 77, and the sum of her scores is 385.385. The desired average is then 80,80, so the sum of scores required is 80⋅6.80\cdot 6. Therefore, the answer is 80⋅6−385=95.80\cdot 6-385 = 95.
Thus, the correct answer is E .
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