2023 AMC 8 Problems/Problem 10


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?

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


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


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.


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

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


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

Video Solution by Magic Square


Video Solution by Interstigation


Video Solution by harungurcan


Video Solution by SpreradTheMathLove


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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