2020 AMC 8 Problems/Problem 2

Revision as of 04:46, 18 November 2020 by Jmansuri (talk | contribs) (Solution 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 $$25$, it follows that he will have to give $$45-$15=$15$ to the others $\implies\boxed{\textbf{(C) }15}$.
~jmansuri

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

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