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Difference between revisions of "1966 AHSME Problems/Problem 1"

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Latest revision as of 11:39, 5 July 2013

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:

$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$

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 • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
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All AHSME Problems and Solutions

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