2020 AMC 8 Problems/Problem 2: Difference between revisions
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The total earnings for the four friends is <math>\$15+\$20+\$25+\$40=\$100</math>. Since they decided to split their earnings equally among themselves, it follows that each person will get <math>\frac{\$100}{4}=\$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=\$15</math> to the others <math>\implies\boxed{\textbf{(C) }\$15}</math>.<br> | The total earnings for the four friends is <math>\$15+\$20+\$25+\$40=\$100</math>. Since they decided to split their earnings equally among themselves, it follows that each person will get <math>\frac{\$100}{4}=\$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=\$15</math> to the others <math>\implies\boxed{\textbf{(C) }\$15}</math>.<br> | ||
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri] | ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri] | ||
==Video Solution== | |||
https://youtu.be/-mSgttsOv2Y | |||
~savannahsolver | |||
==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 11:33, 19 November 2020
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
First we average 
to get 
. Thus, 
. ~~Spaced_Out
Solution 2
The total earnings for the four friends is 
. Since they decided to split their earnings equally among themselves, 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 
.
~junaidmansuri
Video Solution
~savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
