2019 AMC 8 Problems/Problem 11

Revision as of 18:25, 5 July 2020 by Mathnerdnair (talk | contribs) (Solution 3)

Problem 11

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eight graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?

$\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

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

Solution 3

[asy] draw(circle((-0.5,0),1)); draw(circle((0.5,0),1)); label("$\huge{x}$", (0, 0)); label("$70-x$", (-1, 0)); label("$54-x$", (1, 0)); [/asy]

We know that the sum of all three areas is $93$ So, we have: \[93 = 70-x+x+54-x\] \[93 = 70+54-x\] \[93 = 124 - x\] \[-39=-x\] \[x=39\]

We are looking for the number of students in only math. This is $70-x$. Substituting $x$ with $31$, our answer is $\boxed{39}$.

-mathnerdnair

Solution 4

Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA


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

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