2019 AMC 8 Problems/Problem 11

Revision as of 13:35, 18 April 2020 by Happyredpanda (talk | contribs)

Solution 1

Let $x$ be the number of students taking both a math and a foreign language class.

By P-I-E, we get $70 + 54 - x$ = $93$.

Solving gives us $x = 31$.

But we want the number of students taking only a math class.

Which is $70 - 31 = 39$.

$\boxed{\textbf{(D)}\ 39}$

~phoenixfire

Solution 2

We have $70 + 54 = 124$ people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get $70 - 31 = \boxed{\textbf{D} \, 39}$ -fath2012

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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