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

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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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>
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Therefore, the answer is <math>\fbox{D}</math>
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Solution by VivekA
  
 
== See also ==
 
== See also ==

Revision as of 13:41, 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 $\fbox{D}$

Solution by VivekA

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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