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

Revision as of 10:55, 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)
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 ==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>.
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Difference between revisions of "2019 AMC 12A Problems/Problem 1"

Revision as of 10:55, 14 February 2019

Problem

Solution

See Also