Difference between revisions of "2019 AMC 12A Problems/Problem 2"

Revision as of 11:23, 10 February 2019

Problem

Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$ ?

$\textbf{(A) } 50 \qquad \textbf{(B) } 66\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

Solution 1

Since $a=1.5b$ , that means $b=a/1.5$ . We multiply by 3 to get a $3b$ term, to yield $3b=2a$ .

$2a$ is $\boxed{200\%}$ of $a$ .

-- eric2020

Solution 2

WLOG, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\frac{3b}{a}=\frac{300}{150}=2$ so $3b$ is $200\%$ or $a$ so the answer is $\boxed{D}.$

-21jzhang

Solution 3 (Like Solution 1)

$a = 1.5b$ . Multiply by 2 to obtain $2a = 3b$ . Since $2 = 200\%$ , the answer is $\boxed{200\% \implies D}$

-DBlack2021

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 -21jzhang
+==Solution 3 (Like Solution 1)==
+<math>a = 1.5b</math>. Multiply by 2 to obtain <math>2a = 3b</math>. Since <math>2 = 200\%</math>, the answer is <math>\boxed{200\% \implies D}</math>
+-DBlack2021
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