Difference between revisions of "2020 AMC 8 Problems/Problem 2"
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First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = \boxed{\textbf{(C) }15.}</math>. ~~Spaced_Out | First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = \boxed{\textbf{(C) }15.}</math>. ~~Spaced_Out | ||
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==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=1|num-a=3}} | {{AMC8 box|year=2020|num-b=1|num-a=3}} | ||
Revision as of 00:39, 18 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
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. 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
