2020 AMC 8 Problems/Problem 2

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

Solution 2

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

Solution 3

Note that they will each get an equal amount, or the average, so we have $\dfrac{$15+$20+$25+$40}{4}=\dfrac{$100}{4}=$25,$ and so the person with $$40$ will have to give $$40-$25=\boxed{\textbf{(C) }$15}$ to the others.

[pog]

Solution 4

Notice that the friends have $$15+$20+$25+$40=$100$ combined. Hence, they should each have $\frac{$100}{4}=$25$ if they are to slit the bounty equally. The answer then is $$40-x=$25 \Rightarrow x=\textbf{(C)}\ $15$ .

-franzliszt

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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