Art of Problem Solving

2020 AMC 8 Problems/Problem 2: Difference between revisions

Sevenoptimus (talk | contribs)
m Fixed page styling
Sevenoptimus (talk | contribs)
m Minor improvement
Line 5: Line 5:


==Solution==
==Solution==
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>\$40-\$25=\boxed{\textbf{(C) }\$15}</math> to the others.
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.


==Video Solution==
==Video Solution==

Revision as of 09:55, 20 November 2020

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/-mSgttsOv2Y

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing