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2020 AMC 8 Problems/Problem 2: Difference between revisions

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==Problem==
==Problem==
Four friends do yardwork for their neighbors over the weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned <math>\$40</math> give to the others?
Four friends paint dog portraits for their neighbors over the weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned <math>\$40</math> give to the others?


<math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math>
<math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math>

Revision as of 18:47, 27 December 2021

Problem

Four friends paint dog portraits for their neighbors over the weekend, earning $\$15, \$20, \$25,$ and $\$40,$ respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned $\$40$ give to the others?

$\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25$

Solution

The friends earn $\$\left(15+20+25+40\right)=\$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $\$\left(\frac{100}{4}\right)=\$25$. Since the friend who earned $\$40$ will need to leave with $\$25$, he will have to give $\$\left(40-25\right)=\boxed{\textbf{(C) }\$15}$ to the others.

Video Solution by WhyMath

https://youtu.be/-mSgttsOv2Y

~savannahsolver

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=ZwfPEYd55NQ

~North America Math Contest Go Go Go

Video Solution

https://youtu.be/eSxzI8P9_h8

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=62

~Interstigation

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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: Unable to save thumbnail to destination

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