# 2023 AMC 8 Problems/Problem 10

## 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 (HOW TO THINK CREATIVELY!!!)

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