Difference between revisions of "2019 AMC 12A Problems/Problem 2"
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==Solution 2== | ==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{D}.</math> | ||
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==See Also== | ==See Also== | ||
Revision as of 09:09, 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 
-21jzhang
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
