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

Latest revision as of 21:24, 24 December 2023

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

Circumference of a circle is $2 \pi r$ so the radius is $\frac{circumference}{2 \pi}$

So radius of first circle

$2 \pi r = 20$ $r = \frac{10}{\pi}$

Radius of second circle $2 \pi r = 25$ $r = \frac{25}{2 \pi}$

The difference of these radii is

$\frac{25}{2 \pi} - \frac{10}{\pi} = \frac{5}{2 \pi}$

So the answer is $\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$ .

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

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

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

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
+When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
+<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>
+== 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 <math>\pi</math> anymore)
+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>.
+=== Solution 2 ===
+Circumference of a circle is <math>2 \pi r</math> so the radius is <math>\frac{circumference}{2 \pi}</math>
+So radius of first circle
+<cmath>2 \pi r = 20</cmath>
+<cmath>r = \frac{10}{\pi}</cmath>
-When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
+Radius of second circle
+<cmath>2 \pi r = 25</cmath>
+<cmath>r = \frac{25}{2 \pi}</cmath>
+The difference of these radii is
+<cmath>\frac{25}{2 \pi} - \frac{10}{\pi} = \frac{5}{2 \pi}</cmath>
+So the answer is <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.
-<math>\textbf{(A)}\ 5\text{ in} \qquad
+=== Solution 3 ===
-\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad
+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
-\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad
+<cmath>\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}</cmath>
-\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad
+<cmath>4r_2=5r_1</cmath>
-\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math>
+<cmath>4(r_2-r_1)=r_1</cmath>
-==Solution==
+<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}}
-The answer is D.
 ==See Also==
@@ Line 17: / Line 39: @@
 [[Category:Introductory Algebra Problems]]
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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