2019 AMC 12A Problems/Problem 2

Revision as of 12:23, 10 February 2019 by Dblack2021 (talk | contribs) (Added a third solution)

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

See Also

2019 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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. AMC logo.png