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

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==Stupid Dadio.com==
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==The problem!==
 
4 friends do som deed to neighbor over weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. The Friends decide to split earnings equally among themselves. In total how much will friend who earned <math>\$40</math> give to other friends?
 
4 friends do som deed to neighbor over weekend, earning <math>\$15, \$20, \$25,</math> and <math>\$40,</math> respectively. The Friends decide to split earnings equally among themselves. In total how much will friend who earned <math>\$40</math> give to other friends?
 
<math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math>s
 
<math>\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25</math>s

Revision as of 18:30, 19 December 2021

The problem!

4 friends do som deed to neighbor over weekend, earning $$15, $20, $25,$ and $$40,$ respectively. The Friends decide to split earnings equally among themselves. In total how much will friend who earned $$40$ give to other friends? $\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \textbf{(D) }$20 \qquad \textbf{(E) }$25$s

Solution

The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$. Since the friend who earned $$40$ will need to leave with $$25$, he will have to give $$\left(40-25\right)=\boxed{\textbf{(C) }$15}$ to the others.

Video Solution by WhyMath

https://youtu.be/-mSgttsOv2Y

~savannahsolver

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=ZwfPEYd55NQ

~North America Math Contest Go Go Go

Video Solution

https://youtu.be/eSxzI8P9_h8

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=62

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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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