2012 AMC 8 Problems/Problem 4

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

2012 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2012 AMC 8 Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also