1991 AHSME Problems/Problem 16

Problem

One hundred students at Century High School participated in the AHSME last year, and their mean score was 100. The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?

(A) $100$ (B) $112.5$ (C) $120$ (D) $125$ (E) $150$

Solution

Solution by e_power_pi_times_i

Let $s$ and $\dfrac{3}{2}s$ denote the numbers of seniors and non-seniors, respectively. Then $\dfrac{5}{2}s = 100$ , so $s = 40$ , $\dfrac{3}{2}s = 60$ . Let $m$ and $\dfrac{2}{3}m$ denote the mean score of seniors and non-seniors, respectively. Then $40m + 60(\dfrac{2}{3}m) = 40m + 40m = 10000$ . The answer is $\boxed{\textbf{(D) } 125}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1991 AHSME Problems/Problem 16

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