1988 AHSME Problems/Problem 8

Problem

If $\frac{b}{a} = 2$ and $\frac{c}{b} = 3$ , what is the ratio of $a + b$ to $b + c$ ?

$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{3}{8}\qquad \textbf{(C)}\ \frac{3}{5}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{4}$

Solution 1

Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since $b$ is in both equations, that would be a place to start. We manipulate the equations yielding $\frac{b}{2}=a$ and $c=3b$ . Since we are asked to find the ratio of $a+b$ to $b+c$ , we need to find $\frac{a+b}{b+c}$ . We found the $a$ and $c$ in terms of $b$ so that means we can plug them in. We have: $\frac{\frac{b}{2}+b}{b+3b}=\frac{\frac{3}{2}b}{4b}=\frac{3}{8}$ . Thus the answer is $\frac{3}{8} \implies \boxed{\text{B}}$ .

Solution 2

WLOG, let $b=4, a=2, c=12.$ Thus, the answer is $\frac{4+2}{12+4}= \frac{3}{8}$

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1988 AHSME Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See also