Difference between revisions of "2023 AMC 8 Problems/Problem 10"

 
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Harold made a plum pie. Lets call the total amount of this pie <math>x</math>. We can distribute into different fractions of pie that the animals ate. First Harold ate <math>\frac{1}{4} * x</math> of the pie and the remains are <math>\frac{3}{4}</math> of the pie. From this a moose comes and eats <math>\frac{3}{4} * \frac{1}{3}</math> and what remains is <math>\frac{3}{4} - \frac{1}{4} = \frac{1}{2}</math>. From this a porcupine comes and eat <math>\frac{1}{3} * \frac{1}{2} = \frac{1}{6}</math> and our final answer of what remains is <math>\frac{1}{2} - \frac{1}{6} = \boxed{\text{(D)}\frac{1}{3}}</math>
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==Problem==
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Harold made a plum pie to take on a picnic. He was able to eat only <math>\frac{1}{4}</math> of the pie, and he left the rest for his friends. A moose came by and ate <math>\frac{1}{3}</math> of what Harold left behind. After that, a porcupine ate <math>\frac{1}{3}</math> of what the moose left behind. How much of the original pie still remained after the porcupine left?
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<math>\textbf{(A)}\ \frac{1}{12} \qquad \textbf{(B)}\ \frac{1}{6} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{1}{3} \qquad \textbf{(E)}\ \frac{5}{12}</math>
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==Solution==
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Note that:
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* Harold ate <math>\frac14</math> of the pie. After that, <math>1-\frac14=\frac34</math> of the pie was left behind.
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* The moose ate <math>\frac13\cdot\frac34 = \frac14</math> of the pie. After that, <math>\frac34 - \frac14 = \frac12</math> of the pie was left behind.
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* The porcupine ate <math>\frac13\cdot\frac12 = \frac16</math> of the pie. After that, <math>\frac12 - \frac16 = \boxed{\textbf{(D)}\ \frac{1}{3}}</math> of the pie was left behind.
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More simply, we can condense the solution above into the following equation: <cmath>\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.</cmath>
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
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==Remark==
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This problem is a great example of the implementation of <b>remaining portions</b>. This concept, in short, refers to the method of finding out what remains of a whole when part of it is taken. If you look closely at our solution, we can see that the denominator is the same when we said Harold ate <math>\frac14</math> of the pie and we got <math>\frac34</math> of the pie left. We can also see the numerator is just the denominator minus the old denominator.
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~Nivaar
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==Video Solution by Math-X (Let's first Understand the question)==
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https://youtu.be/Ku_c1YHnLt0?si=N3jXnFg5Zy3GCgrB&t=1500 ~Math-X
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==Video Solution (HOW TO THINK CREATIVELY!!!) ==
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https://youtu.be/WKcH7Cyeipo
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~Education the Study of everything
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==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=4814
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=859
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=oIGy79w1H8o&t=246s
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==Video Solution by SpreradTheMathLove==
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https://www.youtube.com/watch?v=TAa6jarbATE
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==Video Solution by Dr. David==
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https://youtu.be/BCEMLHWsw2o
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==Video Solution by WhyMath==
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https://youtu.be/EYCq78Fg1RY
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==See Also==
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{{AMC8 box|year=2023|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 10:09, 18 November 2024

Problem

Harold made a plum pie to take on a picnic. He was able to eat only $\frac{1}{4}$ of the pie, and he left the rest for his friends. A moose came by and ate $\frac{1}{3}$ of what Harold left behind. After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind. How much of the original pie still remained after the porcupine left?

$\textbf{(A)}\ \frac{1}{12} \qquad \textbf{(B)}\ \frac{1}{6} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{1}{3} \qquad \textbf{(E)}\ \frac{5}{12}$

Solution

Note that:

  • Harold ate $\frac14$ of the pie. After that, $1-\frac14=\frac34$ of the pie was left behind.
  • The moose ate $\frac13\cdot\frac34 = \frac14$ of the pie. After that, $\frac34 - \frac14 = \frac12$ of the pie was left behind.
  • The porcupine ate $\frac13\cdot\frac12 = \frac16$ of the pie. After that, $\frac12 - \frac16 = \boxed{\textbf{(D)}\ \frac{1}{3}}$ of the pie was left behind.

More simply, we can condense the solution above into the following equation: \[\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.\]

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM

Remark

This problem is a great example of the implementation of remaining portions. This concept, in short, refers to the method of finding out what remains of a whole when part of it is taken. If you look closely at our solution, we can see that the denominator is the same when we said Harold ate $\frac14$ of the pie and we got $\frac34$ of the pie left. We can also see the numerator is just the denominator minus the old denominator.

~Nivaar

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=N3jXnFg5Zy3GCgrB&t=1500 ~Math-X

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/WKcH7Cyeipo ~Education the Study of everything

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4814

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=859

Video Solution by harungurcan

https://www.youtube.com/watch?v=oIGy79w1H8o&t=246s

Video Solution by SpreradTheMathLove

https://www.youtube.com/watch?v=TAa6jarbATE

Video Solution by Dr. David

https://youtu.be/BCEMLHWsw2o

Video Solution by WhyMath

https://youtu.be/EYCq78Fg1RY

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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