1987 AHSME Problems/Problem 4

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Problem

$\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}$ equals

$\text{(A)} \ 6 \qquad  \text{(B)} \ 8 \qquad  \text{(C)} \ \frac{31}{2} \qquad  \text{(D)} \ 24 \qquad  \text{(E)} \ 512$

Solution

We can factorise to give $\frac{2^1 + 2^0 + 2^{-1}}{2^{-3}(2^1 + 2^0 + 2^{-1})} = \frac{1}{2^{-3}} = 8$, which is $\boxed{\text{B}}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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