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.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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 AMC 12 Problems and Solutions
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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 AMC 10 Problems and Solutions

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