Difference between revisions of "2020 AMC 8 Problems/Problem 2"
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==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>\$\left(40-25\right)=\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. | ||
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| + | ==Video Solution (🚀Just 1 min🚀)== | ||
| + | https://youtu.be/CC5jRJ95QB0 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Revision as of 16:52, 29 July 2023
Contents
Problem
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 (🚀Just 1 min🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
~The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=62
~Interstigation
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=ZwfPEYd55NQ
~North America Math Contest Go Go Go
Video Solution by STEMbreezy
https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=37
~STEMbreezy
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. 
