1966 AHSME Problems/Problem 1

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 $3x - 4 = 3 \Longrightarrow x = \frac 73 \Rightarrow \mathrm{(B)}$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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

Page

Toolbox

Search

1966 AHSME Problems/Problem 1

Problem

Solution

See also