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2009 AMC 10B Problems/Problem 5

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The following problem is from both the 2009 AMC 10B #5 and 2009 AMC 12B #3, so both problems redirect to this page.

Problem

Twenty percent less than 60 is one-third more than what number?

\mathrm{(A)}\ 16\qquad\mathrm{(B)}\ 30\qquad\mathrm{(C)}\ 32\qquad\mathrm{(D)}\ 36\qquad\mathrm{(E)}\ 48

Solution

Twenty percent less than 60 is \frac 45 \cdot 60 = 48. One-third more than a number n is \frac 43n. Therefore \frac 43n = 48 and the number is \boxed {36}. The answer is \mathrm{(D)}.

See also

2009 AMC 10B (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
2009 AMC 12B (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
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