# Difference between revisions of "1950 AHSME Problems/Problem 2"

## Problem

Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$

## Solution

Our first procedure is to find the value of $g$. With the given variables' values, we can see that $8g-4=16$ so $g=\frac{20}{8}=\frac{5}{2}$.

With that, we can replace $g$ with $\frac{5}{2}$. When $S=10$, we can see that $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{\text{(D) 21}}$.