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

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==Solution 2==
 
==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>.
 +
 +
==Solution 3==
 +
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>
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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}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</cmath>
  
 
==Solution 3==
 
==Solution 3==

Revision as of 03:59, 10 June 2020

Problem

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

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

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 $\pi$ 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 \[\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}\] \[4r_2=5r_1\] \[4(r_2-r_1)=r_1\] \[r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}\]

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 \[\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}\] \[4r_2=5r_1\] \[4(r_2-r_1)=r_1\] \[r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\frac{5}{2\pi}\]

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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