Difference between revisions of "2019 AMC 12A Problems/Problem 2"
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<math>\textbf{(A) } 50 \qquad \textbf{(B) } 66\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450</math> | <math>\textbf{(A) } 50 \qquad \textbf{(B) } 66\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450</math> | ||
| − | ==Solution== | + | ==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=a/1.5</math>. We multiply by 3 to get a <math>3b</math> term, to yield <math>3b=2a</math>. | ||
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-- eric2020 | -- eric2020 | ||
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| + | ==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> | ||
==See Also== | ==See Also== | ||
Revision as of 10:09, 10 February 2019
Contents
Problem
Suppose 
is 
of 
. What percent of is 
?
Solution 1
Since 
, that means 
. We multiply by 3 to get a term, to yield 
.
is 
of .
-- eric2020
Solution 2
WLOG, let 
. Then, we have 
and 
. Thus, 
so is 
or so the answer is 
See Also
| 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. 
