Difference between revisions of "2019 AMC 8 Problems/Problem 11"
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==Problem 11== | ==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> | + | 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> eighth 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) } | + | <math>\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70</math> |
==Solution 1== | ==Solution 1== | ||
Line 11: | Line 11: | ||
Solving gives us <math>x = 31</math>. | Solving gives us <math>x = 31</math>. | ||
− | But we want the number of students taking only a math class | + | But we want the number of students taking only a math class, |
− | + | which is <math>70 - 31 = 39</math>. | |
<math>\boxed{\textbf{(D)}\ 39}</math> | <math>\boxed{\textbf{(D)}\ 39}</math> | ||
Line 19: | Line 19: | ||
~phoenixfire | ~phoenixfire | ||
− | ==See | + | ==Solution 2== |
+ | We have <math>70 + 54 = 124</math> 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 <math>31</math> 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 <math>70 - 31 = \boxed{\textbf{D} \, 39}</math>. | ||
+ | |||
+ | -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 <math>93</math> | ||
+ | So, we have: | ||
+ | <cmath>93 = 70-x+x+54-x</cmath> | ||
+ | <cmath>93 = 70+54-x</cmath> | ||
+ | <cmath>93 = 124 - x</cmath> | ||
+ | <cmath>-31=-x</cmath> | ||
+ | <cmath>x=31</cmath> | ||
+ | |||
+ | We are looking for the number of students in only math. This is <math>70-x</math>. Substituting <math>x</math> with <math>31</math>, our answer is <math>\boxed{39}</math>. | ||
+ | |||
+ | -mathnerdnair | ||
+ | |||
+ | ==Solution 4== | ||
+ | We are looking for students in math only, which is the complement (exactly the rest of the students) compared to those taking a language class. Since <math>54</math> students take a language (with or without math), we subtract that from the total number of students. Since <math>93-54= 39,</math> our answer is <math>\boxed{(D) 39}.</math> (It's not necessary to know that <math>70</math> students take math.) | ||
+ | ~hailstone | ||
+ | |||
+ | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
+ | https://youtu.be/IgpayYB48C4?si=hwYx8EF3iCm50Ms8&t=3557 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Solution Explained== | ||
+ | https://youtu.be/gOZOCFNXMhE ~ The Learning Royal | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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 | ||
+ | |||
+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/2K2ZUw-l1ek | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
+ | |||
+ | ==See also== | ||
{{AMC8 box|year=2019|num-b=10|num-a=12}} | {{AMC8 box|year=2019|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Combinatorics Problems]] |
Latest revision as of 20:42, 13 February 2024
Contents
- 1 Problem 11
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution by Math-X (First fully understand the problem!!!)
- 7 Solution Explained
- 8 Solution 5
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution (CREATIVE ANALYSIS!!!)
- 12 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 13 See also
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, we 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
We are looking for students in math only, which is the complement (exactly the rest of the students) compared to those taking a language class. Since students take a language (with or without math), we subtract that from the total number of students. Since
our answer is
(It's not necessary to know that
students take math.)
~hailstone
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=hwYx8EF3iCm50Ms8&t=3557
~Math-X
Solution Explained
https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
Solution 5
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
~savannahsolver
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.