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Difference between revisions of "1950 AHSME Problems/Problem 34"

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==Problem==
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== Problem ==
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When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
  
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When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
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<math>\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math>
  
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<math>\textbf{(A)}\ 5\text{ in} \qquad
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== Solutions ==
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\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad
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=== Solution 1 ===
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\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad
 
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\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad
 
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\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math>
 
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==Solution==
 
 
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore)
 
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore)
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We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>
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We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.
  
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==Solution 2==
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=== Solution 2 ===
 
The radii of the circles are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.
 
The radii of the circles are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.
  
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==Solution 3==
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=== Solution 3 ===
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let the radius of the circle with the larger circumference be <math>r_2</math> and the circle with the smaller circumference be <math>r_1</math>; calculating the ratio of the two
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Let the radius of the circle with the larger circumference be <math>r_2</math> and the circle with the smaller circumference be <math>r_1</math>. Calculating the ratio of the two
 
<cmath>\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}</cmath>
 
<cmath>\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}</cmath>
 
<cmath>4r_2=5r_1</cmath>
 
<cmath>4r_2=5r_1</cmath>
 
<cmath>4(r_2-r_1)=r_1</cmath>
 
<cmath>4(r_2-r_1)=r_1</cmath>
 
<cmath>r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</cmath>
 
<cmath>r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</cmath>
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==Solution 3==
 
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let the radius of the circle with the larger circumference be <math>r_2</math> and the circle with the smaller circumference be <math>r_1</math>; calculating the ratio of the two
 
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<cmath>\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}</cmath>
 
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<cmath>4r_2=5r_1</cmath>
 
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<cmath>4(r_2-r_1)=r_1</cmath>
 
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<cmath>r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\frac{5}{2\pi}</cmath>
 
  
 
==See Also==
 
==See Also==

Revision as of 00:04, 12 October 2020

Problem

When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:

$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$

Solutions

Solution 1

When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be $\pi$ anymore) We see that the circumference was increased by $25\%$. This means the radius was also increased by $25\%$. The radius of the original balloon is $\frac{20}{2\pi}=\frac{10}{\pi}$. With the $25\%$ increase, it becomes $\frac{12.5}{\pi}$. The increase is $\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.

Solution 2

The radii of the circles are $\frac{20}{2\pi}$ and $\frac{25}{2\pi}$, respectively. The positive difference is therefore $\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.

Solution 3

Let the radius of the circle with the larger circumference be $r_2$ and the circle with the smaller circumference be $r_1$. Calculating the ratio of the two \[\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}\] \[4r_2=5r_1\] \[4(r_2-r_1)=r_1\] \[r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}\]

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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All AHSME Problems and Solutions

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