# 1984 AHSME Problems/Problem 6

## Problem

In a certain school, there are $3$ times as many boys as girls and $9$ times as many girls as teachers. Using the letters $b, g, t$ to represent the number of boys, girls, and teachers, respectively, then the total number of boys, girls, and teachers can be represented by the expression $\mathrm{(A) \ }31b \qquad \mathrm{(B) \ }\frac{37b}{27} \qquad \mathrm{(C) \ } 13g \qquad \mathrm{(D) \ }\frac{37g}{27} \qquad \mathrm{(E) \ } \frac{37t}{27}$

## Solution

From the given, we have $3g=b$ and $9t=g$, or $t=\frac{g}{9}$. The sum of these, in terms of $g$, is $3g+g+\frac{g}{9}$, or, with a common denominator, $\frac{37g}{9}$. We can see that this isn't one of the choices. So we write it in terms of $b$. We can see from the first equation that $g=\frac{b}{3}$, so substituting this into the expression yields $\frac{37b}{27}, \boxed{\text{B}}$.

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