Difference between revisions of "1968 AHSME Problems/Problem 1"

Revision as of 12:40, 28 July 2016

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 $\fbox{D}$

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 2
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 == Solution ==
-Let <math>d</math> be the diameter of the original circle. If <math>d</math> is increased by <math>\pi</math>, then the new circumference is <math>\pi d + \pi^2 \Rightarrow \pi d + \pi^2 - \pi d = \pi ^2</math>
+Let <math>d</math> be the diameter of the original circle. If <math>d</math> is increased by <math>\pi</math>, then the new circumference is <math>\pi d + \pi^2</math>. The difference in circumference is therefore <math>\pi d + \pi^2 - \pi d = \pi^2</math>
-<math>\fbox{D}</math>
+Therefore, the answer is <math>\fbox{D}</math>
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Difference between revisions of "1968 AHSME Problems/Problem 1"

Revision as of 12:40, 28 July 2016

Problem

Solution

See also