Difference between revisions of "2020 AMC 8 Problems/Problem 2"

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First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = 15 \implies \boxed{C}</math>. - Spaced_Out
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==Problem==
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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?
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<math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math>
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==Solution==
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The friends earn <math>\$\left(15+20+25+40\right)=\$100</math> in total. Since they decided to split their earnings equally, it follows that each person will get <math>\$\left(\frac{100}{4}\right)=\$25</math>. Since the friend who earned <math>\$40</math> will need to leave with <math>\$25</math>, he will have to give <math>\$\left(40-25\right)=\boxed{\textbf{(C) }\$15}</math> to the others.
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==Video Solution==
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https://youtu.be/eSxzI8P9_h8
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==See also==
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{{AMC8 box|year=2020|num-b=1|num-a=3}}
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{{MAA Notice}}

Revision as of 00:21, 10 January 2021

Problem

Four friends do yardwork 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

https://youtu.be/eSxzI8P9_h8

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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

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