Difference between revisions of "2020 AMC 8 Problems/Problem 2"
Spaced out (talk | contribs) (Created page with "First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = 15 \implies \boxed{C}</math>. - Spaced_Out") |
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| − | First we average <math>15,20,25,40</math> to get <math>25</math>. Thus, <math>40 - 25 = | + | ==Problem 2== |
| + | Four friends do yardwork for their neighbors over the weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned <math>\$40</math> give to the others? | ||
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| + | <math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math> | ||
| + | |||
| + | ==Solution== | ||
| + | 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 | ||
Revision as of 23:55, 17 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
