1968 AHSME Problems/Problem 1

Problem

Let $P$ units be the increase in circumference of a circle resulting from an increase in $\pi$ units in the diameter. Then $P$ equals:

$\text{(A) } \frac{1}{\pi}\quad\text{(B) } \pi\quad\text{(C) } \frac{\pi^2}{2}\quad\text{(D) } \pi^2\quad\text{(E) } 2\pi$

Solution

Let $d$ be the diameter of the original circle. If $d$ is increased by $\pi$ , then the new circumference is $\pi d + \pi^2$ . The difference in circumference is therefore $\pi d + \pi^2 - \pi d = \pi^2$

Therefore, the answer is $\fbox{D}$

Solution by VivekA

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 2
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1968 AHSME Problems/Problem 1

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