Difference between revisions of "2020 AMC 8 Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
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> | ||
| − | ~ junaidmansuri | + | ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri] |
==Solution 3== | ==Solution 3== | ||
Revision as of 18:31, 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
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
Solution 3
Note that they will each get an equal amount, or the average, so we have 
and so the person with will have to give 
to the others.
[pog]
Solution 4
Notice that the friends have combined. Hence, they should each have if they are to slit the bounty equally. The answer then is 
.
-franzliszt
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. 
