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 == | ||
| − | <math>\ | + | Solution by e_power_pi_times_i |
| + | |||
| + | |||
| + | 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>. | ||
== See also == | == See also == | ||
Latest revision as of 11:47, 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 
more than the number of seniors, and the mean score of the seniors was higher than that of the non-seniors. What was the mean score of the seniors?
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
Solution by e_power_pi_times_i
Let 
and 
denote the numbers of seniors and non-seniors, respectively. Then 
, so 
, 
. Let 
and 
denote the mean score of seniors and non-seniors, respectively. Then 
. The answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
