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2020 AMC 8 Problems/Problem 5

Problem

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?

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

Solution 1

Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = \frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{\textbf{(C) }15}$.

Solution 2

The pitcher is $\frac{3}{4}$ full, i.e. $75\%$ full. Therefore each cup receives $\frac{75}{5}=\boxed{\textbf{(C) }15}$ percent of the total capacity.

Solution 3

Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that each cup received $15\%$ of the total capacity of the pitcher, yielding $\boxed{\textbf{(C) }15}$ as the answer.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=dBVy_7cb8HQGTdBs&t=433

~Math-X

Video Solution (🚀Very Fast🚀)

https://youtu.be/_6cEquyjWXQ

~Education, the Study of Everything

Video Solution by North America Math Contest go go go

https://www.youtube.com/watch?v=bRzVP_caKrU ~North America Math Contest go go go

Video Solution by WhyMath

https://youtu.be/ph_qAhXXKP4

~savannahsolver

Video Solution

https://youtu.be/eSxzI8P9_h8 ~ The Learning Royal

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=165

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=205

~STEMbreezy

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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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