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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==Solution 1==
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Harold made a plum pie. We can distribute into different fractions of pie that the animals ate. First Harold ate <math>\frac{1}{4}</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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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Revision as of 18:48, 24 January 2023

Solution 1

Harold made a plum pie. We can distribute into different fractions of pie that the animals ate. First Harold ate $\frac{1}{4}$ of the pie and the remains are $\frac{3}{4}$ of the pie. From this a moose comes and eats $\frac{3}{4} * \frac{1}{3}$ and what remains is $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$. From this a porcupine comes and eat $\frac{1}{3} * \frac{1}{2} = \frac{1}{6}$ and our final answer of what remains is $\frac{1}{2} - \frac{1}{6} =  \boxed{\text{(D)}\frac{1}{3}}$.


~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat