Difference between revisions of "2019 AMC 8 Problems/Problem 11"
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==Solution 4== | ==Solution 4== | ||
Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA | Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA | ||
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+ | == Video Solution == | ||
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+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Kanl4ni2y0o&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=12 | ||
==See also== | ==See also== |
Revision as of 13:50, 23 April 2021
Contents
Problem 11
The eighth grade class at Lincoln Middle School has students. Each student takes a math class or a foreign language class or both. There are eighth graders taking a math class, and there are eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?
Solution 1
Let be the number of students taking both a math and a foreign language class.
By P-I-E, we get = .
Solving gives us .
But we want the number of students taking only a math class.
Which is .
~phoenixfire
Solution 2
We have 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 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 -fath2012
Solution 3
We know that the sum of all three areas is So, we have:
We are looking for the number of students in only math. This is . Substituting with , our answer is .
-mathnerdnair
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.