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

Revision as of 20:48, 17 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 $b=\frac{a}{1.5}$ . 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 $\boxed{\textbf{(D) }200\%}$ .

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

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

@@ Line 6: / Line 6: @@
 ==Solution 1==
-Since <math>a=1.5b</math>, that means <math>b=a/1.5</math>. We  multiply by 3 to get a <math>3b</math> term, to yield <math>3b=2a</math>.
+Since <math>a=1.5b</math>, that means <math>b=\frac{a}{1.5}</math>. We multiply by <math>3</math> to get a <math>3b</math> term, yielding <math>3b=2a</math>, and <math>2a</math> is <math>\boxed{\textbf{(D) }200\%}</math> of <math>a</math>.
-<math>2a</math> is <math>\boxed{\textbf{(D) }200\%}</math> of <math>a</math>.
--- eric2020
 ==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{\textbf{(D) }200\%}</math>.
+Without loss of generality, 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> of <math>a</math>. Hence the answer is <math>\boxed{\textbf{(D) }200\%}</math>.
--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{\textbf{(D) }200\%}</math>.
--DBlack2021
+==Solution 3 (similar to Solution 1)==
+As before, <math>a = 1.5b</math>. Multiply by 2 to obtain <math>2a = 3b</math>. Since <math>2 = 200\%</math>, the answer is <math>\boxed{\textbf{(D) }200\%}</math>.
-==Solution 4 (Like Solution 2)==
+==Solution 4 (similar to Solution 2)==
-WLOG, let <math>b=2</math>. Then, we have <math>a=3</math> and <math>3b=6</math>.  Thus, <math>\frac{3b}{a}=\frac{6}{3}=2</math> so <math>3b</math> is <math>200\%</math> of <math>a</math> so the answer is <math>\boxed{\textbf{(D) }200\%}</math>.
+Without loss of generality, let <math>b=2</math>. Then, we have <math>a=3</math> and <math>3b=6</math>. This gives <math>\frac{3b}{a}=\frac{6}{3}=2</math>, so <math>3b</math> is <math>200\%</math> of <math>a</math>, so the answer is <math>\boxed{\textbf{(D) }200\%}</math>.
 ==See Also==

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