Difference between revisions of "2019 AMC 12A Problems/Problem 1"

Revision as of 11:55, 14 February 2019

Problem

The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$ ?

$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$

Solution

The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{\textbf{(E) }78}$ .

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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
All AMC 12 Problems and Solutions

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

@@ Line 7: / Line 7: @@
 ==Solution==
-Let the first jar's volume be <math>A</math> and the second's be <math>B</math>. It is given that <math>\frac{3}{4}A=\frac{5}{6}B</math>. We find that <math>\frac{B}{A}=\frac{3/4}{5/6}=\boxed{\frac{9}{10}}.</math> We know that this is the ratio of smaller to larger volume already because it is less than <math>1.</math>
+The area of the larger pizza is <math>16\pi</math>, while the area of the smaller pizza is <math>9\pi</math>. Therefore, the larger pizza is <math>\frac{7\pi}{9\pi} \cdot 100\%</math> bigger than the smaller pizza. <math>\frac{7\pi}{9\pi} \cdot 100\% = 77.777....</math>, which is closest to <math>\boxed{\textbf{(E) }78}</math>.
 ==See Also==

Page

Toolbox

Search

Difference between revisions of "2019 AMC 12A Problems/Problem 1"

Revision as of 11:55, 14 February 2019

Problem

Solution

See Also