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Difference between revisions of "2020 AMC 8 Problems/Problem 2"

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==Solution==
 
==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
 
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
 +
 +
==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{</math>100}{4}=\$25<math>. Since the friend who earned </math>\$40<math> will need to leave with </math>\$25<math>, it follows that he will have to give </math>45-15=\$15<math> to the others </math>\implies\boxed{\textbf{(C) }15}$.<br>
 +
~jmansuri
  
 
==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 03:45, 18 November 2020

Problem 2

Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned $$40$ give to the others?

$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \textbf{(D) }$20 \qquad \textbf{(E) }$25$

Solution

First we average $15,20,25,40$ to get $25$. Thus, $40 - 25 = \boxed{\textbf{(C) }15.}$. ~~Spaced_Out

Solution 2

The total earnings for the four friends is $15+20+25+40=$100$. Since they decided to split their earnings equally among themselves, it follows that each person will get $\frac{$ (Error compiling LaTeX. Unknown error_msg)100}{4}=$25$. Since the friend who earned$$40$will need to leave with$$25$, it follows that he will have to give$45-15=$15$to the others$\implies\boxed{\textbf{(C) }15}$.
~jmansuri

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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