Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4)
Line 48: Line 48:
 
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.)
 
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
 
~hailstone
 +
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=hwYx8EF3iCm50Ms8&t=3557
 +
 +
~Math-X
  
 
==Solution Explained==
 
==Solution Explained==

Revision as of 15:32, 30 December 2023

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 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 $54$ students take a language (with or without math), we subtract that from the total number of students. Since $93-54= 39,$ our answer is $\boxed{(D) 39}.$ (It's not necessary to know that $70$ 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&oldid=208976"