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

Latest revision as of 17:31, 30 October 2022

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 $\boxed{\textbf{(D) }200\%}$ .

Video Solution 1

https://youtu.be/CM2PqoCDvqo

~Education, the Study of Everything

2019 AMC 12A (Problems • Answer Key • Resources)
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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 16: / Line 16: @@
 ==Solution 4 (similar to Solution 2)==
 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>.
+==Video Solution 1==
+https://youtu.be/CM2PqoCDvqo
+~Education, the Study of Everything
 ==See Also==

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