Difference between revisions of "2012 AMC 8 Problems/Problem 4"

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==Solution 2==
 
==Solution 2==
An alternative way of solving this problem is to add the slices separately. When Peter takes a full slice, he takes <math>\frac{1}{12}</math> of the pizza. When he shares the slice with paul, he splits a slice of pizza into two, the equation is <math>\frac{1}{2}</math> of <math>\frac{1}{12}</math>, which is $\frac{1}{24}. Adding gives: <cmath>\frac{1}{12}+\frac{1}{24}=\boxed{\textbf{(C)}~\frac{1}{8}}</cmath>
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An alternative way of solving this problem is to add the slices separately. When Peter takes a full slice, he takes <math>\frac{1}{12}</math> of the pizza. When he shares the slice with paul, he splits a slice of pizza into two, the equation is <math>\frac{1}{2}</math> of <math>\frac{1}{12}</math>, which is <math>\frac{1}{24}</math>. Adding gives: <cmath>\frac{1}{12}+\frac{1}{24}=\boxed{\textbf{(C)}~\frac{1}{8}}</cmath>
  
 
~SmartGrowth ~edited by megaboy6679
 
~SmartGrowth ~edited by megaboy6679

Latest revision as of 00:14, 29 January 2023

Problem

Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?

$\textbf{(A)}~\frac{1}{24}\qquad\textbf{(B)}~\frac{1}{12}\qquad\textbf{(C)}~\frac{1}{8}\qquad\textbf{(D)}~\frac{1}{6}\qquad\textbf{(E)}~\frac{1}{4}$

Solution 1

Peter ate $1 + \frac{1}{2} = \frac{3}{2}$ slices. The pizza has $12$ slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate $\dfrac{\frac{3}{2}\text{ slices}}{12\text{ slices}} = \boxed{\textbf{(C)}\ \frac{1}{8}}$ of the pizza.

Solution 2

An alternative way of solving this problem is to add the slices separately. When Peter takes a full slice, he takes $\frac{1}{12}$ of the pizza. When he shares the slice with paul, he splits a slice of pizza into two, the equation is $\frac{1}{2}$ of $\frac{1}{12}$, which is $\frac{1}{24}$. Adding gives: \[\frac{1}{12}+\frac{1}{24}=\boxed{\textbf{(C)}~\frac{1}{8}}\]

~SmartGrowth ~edited by megaboy6679

Video Solution

https://youtu.be/WDRPwoRbpKo ~savannahsolver

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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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