# 2019 AMC 12A Problems/Problem 2

## 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 $b=\frac{a}{1.5}$. We multiply by $3$ to get a $3b$ term, yielding $3b=2a$, and $2a$ is $\boxed{\textbf{(D) }200\%}$ of $a$.

## Solution 2

Without loss of generality, let $b=100$. Then, we have $a=150$ and $3b=300$. Thus, $\frac{3b}{a}=\frac{300}{150}=2$, so $3b$ is $200\%$ of $a$. Hence the answer is $\boxed{\textbf{(D) }200\%}$.

## Solution 3 (similar to Solution 1)

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

## Solution 4 (similar to Solution 2)

Without loss of generality, let $b=2$. Then, we have $a=3$ and $3b=6$. This gives $\frac{3b}{a}=\frac{6}{3}=2$, so $3b$ is $200\%$ of $a$, so the answer is $\boxed{\textbf{(D) }200\%}$.

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