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Difference between revisions of "1991 AHSME Problems/Problem 16"
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== Problem ==
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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 <math>50\%</math> more than the number of seniors, and the mean score of the seniors was <math>50\%</math> higher than that of the non-seniors. What was the mean score of the seniors?
 
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 <math>50\%</math> more than the number of seniors, and the mean score of the seniors was <math>50\%</math> higher than that of the non-seniors. What was the mean score of the seniors?
  
 
(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>
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== Solution ==
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Solution by e_power_pi_times_i
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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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== See also ==
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{{AHSME box|year=1991|num-b=15|num-a=17}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:47, 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

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

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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