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

Revision as of 22:37, 14 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 $\boxed{\textbf{(D) }200\%}$ .

-21jzhang

Solution 3 (Like Solution 1)

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

-DBlack2021

Solution 4 (Like Solution 2)

WLOG, let $b=2$ . Then, we have $a=3$ and $3b=6$ . Thus, $\frac{3b}{a}=\frac{6}{3}=2$ so $3b$ is $200\%$ of $a$ so the answer is $\boxed{\textbf{(D) }200\%}$ .

2019 AMC 12A (Problems • Answer Key • Resources)
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.

@@ Line 13: / Line 13: @@
 ==Solution 2==
-WLOG, let <math>b=100</math>. Then, we have <math>a=150</math> and <math>3b=300</math>.  Thus, <math>\frac{3b}{a}=\frac{300}{150}=2</math> so <math>3b</math> is <math>200\%</math> or <math>a</math> so the answer is <math>\boxed{D}.</math>
+WLOG, let <math>b=100</math>. Then, we have <math>a=150</math> and <math>3b=300</math>.  Thus, <math>\frac{3b}{a}=\frac{300}{150}=2</math> so <math>3b</math> is <math>200\%</math> or <math>a</math> so the answer is <math>\boxed{\textbf{(D) }200\%}</math>.
 -21jzhang

Page

Toolbox

Search