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

1966 AMC 12 Problem 1

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Given that the ratio of

3x-4

y+15

is constant, and

y=3

when

x=2,

then, when

y=12,

x

equals:

\dfrac18

\dfrac87

\dfrac73

\dfrac72

8

Use

x=2

and

y=3

to find the constant ratio

Keep

\frac{3x-4}{y+15}

equal to that ratio when

y=12

The given pair makes the ratio

\frac{3\cdot2-4}{3+15}=\frac{1}{9}.

Therefore, when

y=12,

\frac{3x-4}{27}=\frac19,

3x-4=3

and

x=\frac{7}{3}.

Therefore, the correct answer is C.