2019 AMC 8 Problems/Problem 11: Difference between revisions

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Revision as of 12:35, 18 April 2020

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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-==Problem 11==
-The eighth grade class at Lincoln Middle School has <math>93</math> students. Each student takes a math class or a foreign language class or both. There are <math>70</math> eighth graders taking a math class, and there are <math>54</math> eight graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?
-<math>\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70</math>
 ==Solution 1==
 Let <math>x</math> be the number of students taking both a math and a foreign language class.

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2019 AMC 8 Problems/Problem 11: Difference between revisions

Revision as of 12:35, 18 April 2020

Solution 1

Solution 2

See Also