1981 AHSME Problems/Problem 4

Problem

If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is:

$\text{(A)}\quad 16 \qquad \text{(B)}\quad 24 \qquad \text{(C)}\quad 32 \qquad \text{(D)}\quad 44 \qquad \text{(E)} \quad 52$

Let the smaller number be $x$ and the larger number be $y.$ We can use the information given in the problem to write two equations:

1) Three times the larger of two numbers is four times the smaller. $3y = 4x$

2) The difference between the number is 8.

$y - x = 8$

Solving this system of equation yields $x = 24$ and $y = 32.$ The answer is $\text{C}.$