1992 AHSME Problems/Problem 1

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Problem

If $3(4x+5\pi)=P$ then $6(8x+10\pi)=$

$\text{(A) } 2P\quad \text{(B) } 4P\quad \text{(C) } 6P\quad \text{(D) } 8P\quad \text{(E) } 18P$

Solution

We can see that $8x+10\pi$ is equal to $2(4x+5\pi),$ and we know that $2^2 = 4,$ so the answer is $\boxed{B}\, .$

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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