# Difference between revisions of "1990 AHSME Problems/Problem 24"

## Problem

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?

$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{Average Scores}\\\hline Category&Adams&Baker&Adams\&Baker\\\hline Boys&71&81&79\\ Girls&76&90&?\\ Boys\&Girls&74&84& \\\hline \end{tabular}$

$\text{(A) } 81\quad \text{(B) } 82\quad \text{(C) } 83\quad \text{(D) } 84\quad \text{(E) } 85$

## Solution

Let the numbers of boys and girls at Adams be $(A,a)$ and the numbers of boys and girls at Baker be $(B,b)$.

Then, reading down, we have $71A+76a=74(A+a)$ and $81B+90b=84(B+b)$. Reading across, we have $71A+81B=79(A+B)$.

Simplifying each in turn, $3A-2a=0 \implies A=2a/3$ and $3B-6b=0 \implies B=2b$ and $8A-2B=0 \implies 4A=B$.

Thus $3b = 4a$, which implies that Adams has $\tfrac37$ of the girls and Baker has $\tfrac47$ of the girls.

So the average score for all girls must be $\frac{3\cdot 76+4\cdot 90}7 = \frac{228+360}7 = 84$ which is $\fbox{D}$