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:

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

Solution

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 $\text{C}.$

See also

1981 AHSME (ProblemsAnswer KeyResources)
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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