Difference between revisions of "2008 AMC 12A Problems/Problem 1"
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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>. | 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>. | ||
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+ | 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 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}} |
Latest revision as of 17:18, 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.