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{{MAA Notice}}

Latest revision as of 15:03, 22 April 2021

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 (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions
2009 AMC 12B (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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

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