Difference between revisions of "2020 AMC 8 Problems/Problem 2"
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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] | ||
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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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:53, 18 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
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
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 |
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