Difference between revisions of "2019 AMC 8 Problems/Problem 11"

Revision as of 15:56, 19 August 2020

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 Video Solution - https://youtu.be/Lw8fSbX_8FU, (also explains problems 11-20)

@@ Line 45: / Line 45: @@
 ==Solution 4==
 Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA
+Video Solution - https://youtu.be/Lw8fSbX_8FU, (also explains problems 11-20)
 ==See Also==

2019 AMC 8 (Problems • Answer Key • Resources)
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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