1990 AMC 12 Problem 24

Attempt Problem 24 of the 1990 AMC 12 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 1990 AMC 12 solutions, or check the answer key.

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

24.

All students at Adams High School and at Baker High School take a certain exam. The average scores for boys, for girls, and for boys and girls combined, at Adams HS and Baker HS are shown in the table, as is the average for boys at the two schools combined. What is the average score for the girls at the two schools combined?
Adams Baker Adams & Baker
Boys: 7171 8181 7979
Girls: 7676 9090 ?
Boys & Girls: 7474 8484
8181
8282
8383
8484
8585
Answer: D
Concepts:weighted averageratiosystem of equations
Small hint:
Use each school’s combined average to find its ratio of girls to boys
Big hint:
Use the combined boys’ average to relate the numbers of boys at the two schools
Solution:
Let Adams have xx boys and yy girls. From its average, 71x+76y=74(x+y), 71x+76y=74(x+y), so y=3x2.y=\frac{3x}{2}. If Baker has uu boys and vv girls, its average gives u=2v.u=2v. The combined boys’ average gives 71x+81u=79(x+u), 71x+81u=79(x+u), so u=4xu=4x and v=2x.v=2x. Hence the combined girls’ average is 76(3x2)+90(2x)3x2+2x=84. \frac{76(\frac{3x}{2})+90(2x)}{\frac{3x}{2}+2x}=84.
Thus the correct answer is D.
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