Art of Problem Solving
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

1950 AHSME Problems/Problem 8: Difference between revisions

← Older edit
Samma (talk | contribs)
editor
9 edits
Samma (talk | contribs)
editor
9 edits
 
Line 10: Line 10:


Since the area of the circle is given by the formula <imath>\pi r^2,</imath> the area of the new circle is <imath>\pi (2r)^2 = 4\pi r^2.</imath> The area is quadrupled, or increased by <imath>\boxed{\mathrm{(C) }300\%.}</imath>
Since the area of the circle is given by the formula <imath>\pi r^2,</imath> the area of the new circle is <imath>\pi (2r)^2 = 4\pi r^2.</imath> The area is quadrupled, or increased by <imath>\boxed{\mathrm{(C) }300\%.}</imath>
==Solution==
\[
\textbf{Alternate Solution:}
\]
Let the original radius be \( r \) and the original area be
\[
A_1 = \pi r^2.
\]
A \(100\%\) increase in radius means the new radius is
\[
r_2 = r + 100\% \times r = 2r.
\]
The new area is
\[
A_2 = \pi (r_2)^2 = \pi (2r)^2 = 4\pi r^2.
\]
Now, compute the percent increase in area:
\[
\text{Percent Increase} = \frac{A_2 - A_1}{A_1} \times 100\%
= \frac{4\pi r^2 - \pi r^2}{\pi r^2} \times 100\%
= 3 \times 100\% = 300\%.
\]
\[
\boxed{\text{The area is increased by } 300\%.}
\]


==See Also==
==See Also==

Latest revision as of 17:45, 9 November 2025

Problem

If the radius of a circle is increased $100\%$, the area is increased:

$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$

Solution

Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$

Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{\mathrm{(C) }300\%.}$

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1950_AHSME_Problems/Problem_8&oldid=268122"