Difference between revisions of "2008 AMC 12A Problems/Problem 1"
I like pie (talk | contribs) m (Typo) |
(→See Also) |
||
| Line 13: | Line 13: | ||
{{AMC12 box|year=2008|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2008|ab=A|before=First Question|num-a=2}} | ||
{{AMC10 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}} | ||
Revision as of 20:34, 3 July 2013
- 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 
.
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. 
