Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==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{ ...")
 
(→‎Solution 2)
 
(11 intermediate revisions by 8 users not shown)
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>
 +
 
 +
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>.
  
−
When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:
+
=== 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>
 +
<cmath>4r_2=5r_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>
 +
 
 +
==See Also==
 +
{{AHSME 50p box|year=1950|num-b=33|num-a=35}}
  
−
<math>\textbf{(A)}\ 5\text{ in} \qquad
+
[[Category:Introductory Algebra Problems]]
−
\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad
+
[[Category:Introductory Geometry Problems]]
−
\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad
+
{{MAA Notice}}
−
\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad
 
−
\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math>
 

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_34&oldid=208338"