2009 AMC 10B Problems/Problem 5

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 $\mathrm{(D)}$ .

2009 AMC 10B (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

2009 AMC 12B (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2009 AMC 10B Problems/Problem 5

Problem

Solution

See also