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Difference between revisions of "1991 AHSME Problems/Problem 16"

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(A) <math>100</math> (B) <math>112.5</math> (C) <math>120</math> (D) <math>125</math> (E) <math>150</math>
 
(A) <math>100</math> (B) <math>112.5</math> (C) <math>120</math> (D) <math>125</math> (E) <math>150</math>
 
== Solution ==
 
== Solution ==
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Let <math>s</math> and <math>\dfrac{3}{2}s</math> denote the numbers of seniors and non-seniors, respectively. Then <math>\dfrac{5}{2}s = 100</math>, so <math>s = 40</math>, <math>\dfrac{3}{2}s = 60</math>. Let <math>m</math> and <math>\dfrac{2}{3}m</math> denote the mean score of seniors and non-seniors, respectively. Then <math>40m + 60(\dfrac{2}{3}m) = 40m + 40m = 10000</math>. The answer is <math>\boxed{\textbf{(D) } 125}</math>.
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<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
  

Revision as of 11:44, 13 December 2016

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

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 $\boxed{\textbf{(D) } 125}$.

$\fbox{D}$

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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