Difference between revisions of "2008 AMC 12A Problems/Problem 1"
(→Solution) |
|||
| (8 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #1]] and [[2008 AMC 10A Problems/Problem 1|2008 AMC 10A #1]]}} | ||
==Problem == | ==Problem == | ||
| − | A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day's job. At what time will the doughnut machine complete the job? | + | A bakery owner turns on his doughnut machine at <math>\text{8:30}\ {\small\text{AM}}</math>. At <math>\text{11:10}\ {\small\text{AM}}</math> the machine has completed one third of the day's job. At what time will the doughnut machine complete the job? |
| − | <math>\ | + | <math>\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}}</math> |
==Solution== | ==Solution== | ||
| − | The machine completes one third of the job in <math>\text{11:10}-\text{8:30}=\text{2:40}</math>. Thus the entire job is completed in <math>3 \cdot (\text{2:40}) = \text{8:00}</math>. Since the machine | + | The machine completes one-third of the job in <math>\text{11:10}-\text{8:30}=\text{2:40}</math> hours. Thus, the entire job is completed in <math>3\cdot(\text{2:40})=\text{8:00}</math> hours. |
| + | |||
| + | Since the machine was started at <math>\text{8:30 AM}</math>, the job will be finished <math>8</math> hours later, at <math>\text{4:30 PM}</math>. The answer is <math>\mathrm{(D)}</math>. | ||
| + | |||
| + | Note: <math>\text{2:40}</math> means <math>2</math> hours and <math>40</math> minutes. <math>3</math> multiplied by this time interval is <math>8</math> hours. | ||
==See Also== | ==See Also== | ||
| − | {{AMC12 box|year=2008|ab=A|before=First | + | {{AMC12 box|year=2008|ab=A|before=First Question|num-a=2}} |
| + | {{AMC10 box|year=2008|ab=A|before=First Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:18, 4 June 2021
- 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 
. At 
the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?
Solution
The machine completes one-third of the job in 
hours. Thus, the entire job is completed in 
hours.
Since the machine was started at 
, the job will be finished 
hours later, at 
. The answer is 
.
Note: 
means 
hours and 
minutes. 
multiplied by this time interval is hours.
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
