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 2==
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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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First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = \boxed{\textbf{(C) }15.}</math>. ~~Spaced_Out

Revision as of 00:55, 18 November 2020

Problem 2

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

First we average $15,20,25,40$ to get $25$. Thus, $40 - 25 = \boxed{\textbf{(C) }15.}$. ~~Spaced_Out