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

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

Solution

Let $x$ be the smaller number and $y$ be the larger number. Then $3y = 4x$ , so $x = \frac{3}{4}y$ . Also, $y - x = 8$ , so $y - \frac{3}{4}y = \frac{1}{4}y = 8$ , and $y = \boxed{(\textbf{C})\ 32}$ .

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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1981 AHSME Problems/Problem 4

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