# Difference between revisions of "2008 AMC 12A Problems/Problem 1"

The following problem is from both the 2008 AMC 12A #1 and 2008 AMC 10A #1, so both problems redirect to this page.

## Problem

A bakery owner turns on his doughnut machine at $\text{8:30}\ {\small\text{AM}}$. At $\text{11:10}\ {\small\text{AM}}$ the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?

$\mathrm{(A)}\ \text{1:50}\ {\small\text{PM}}\qquad\mathrm{(B)}\ \text{3:00}\ {\small\text{PM}}\qquad\mathrm{(C)}\ \text{3:30}\ {\small\text{PM}}\qquad\mathrm{(D)}\ \text{4:30}\ {\small\text{PM}}\qquad\mathrm{(E)}\ \text{5:50}\ {\small\text{PM}}$

## Solution

The machine completes one-third of the job in $\text{11:10}-\text{8:30}=\text{2:40}$ hours. Thus, the entire job is completed in $3\cdot(\text{2:40})=\text{8:00}$ hours.

Since the machine was started at $\text{8:30 AM}$, the job will be finished $8$ hours later, at $\text{4:30 PM}$. The answer is $\mathrm{(D)}$.

Note: $\text{2:40}$ means $2$ hours and $40$ minutes. $3$ multiplied by this time interval is $8$ hours.