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1966 AHSME Problems/Problem 1

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Problem

Given that the ratio of 3x - 4 to y + 15 is constant, and y = 3 when x = 2, then, when y = 12, x equals:

\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} ...

Solution

Let k be the constant ratio. Then k = \frac{3(2)-4}{(3)+15} = \frac{1}{9} = \frac{3x - 4}{(12) + 15}. Solving gives 3x - 4 = 3 \Longrightarrow x = \frac 73 \Rightarrow \mathrm{(B)}.

See also

1966 AHSME (Problems)
Preceded by
First question
Followed by
Problem 2
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