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

Revision as of 10:34, 26 January 2022

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 $54$ eighth 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 we 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: $93 = 70-x+x+54-x$ $93 = 70+54-x$ $93 = 124 - x$ $-31=-x$ $x=31$

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 Explained

https://youtu.be/gOZOCFNXMhE ~ The Learning Royal

Solution 4

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

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Kanl4ni2y0o&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=12

Video Solution

https://youtu.be/4E8I-NJYAJY

~savannahsolver

2019 AMC 8 (Problems • Answer Key • Resources)
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Difference between revisions of "2019 AMC 8 Problems/Problem 11"

Revision as of 10:34, 26 January 2022

Contents

Problem 11

Solution 1

Solution 2

Solution 3

Solution Explained

Solution 4

Video Solution

Video Solution

See also

@@ Line 52: / Line 52: @@
 Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Kanl4ni2y0o&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=12
+==Video Solution==
+https://youtu.be/4E8I-NJYAJY
+~savannahsolver
 ==See also==