Difference between revisions of "2020 AMC 8 Problems/Problem 2"
Sevenoptimus (talk | contribs) (Removed solution 2 (effectively the same as solution 1), fixed LaTeX and grammar etc.) |
|||
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. | |
− | |||
− | |||
− | |||
− | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/-mSgttsOv2Y | https://youtu.be/-mSgttsOv2Y | ||
− | |||
− | |||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=1|num-a=3}} | {{AMC8 box|year=2020|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:54, 20 November 2020
Contents
Problem 2
Four friends do yardwork for their neighbors over the weekend, earning and respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned give to the others?
Solution
The friends earn in total. Since they decided to split their earnings equally, it follows that each person will get . Since the friend who earned will need to leave with , he will have to give to the others.
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
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.