Difference between revisions of "2008 AMC 12A Problems/Problem 1"
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| + | ==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? | ||
| + | <math>\textbf{(A)} \text{ 1:50 PM } \qquad \textbf{(B)} \text{ 3:00 PM } \qquad \textbf{(C)} \text{ 3:30 PM } \qquad \textbf{(D)} \text{ 4:30 PM } \qquad \textbf{(E)} \text{ 5:50 PM }</math> | ||
| + | |||
| + | ==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 is started at <math>\text{8:30 AM}</math>, the job is finished <math>8</math> hours later at <math>\text{4:30 PM}\Rightarrow D</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2008|ab=A|before=First question|num-a=2}} | ||
Revision as of 13:41, 17 February 2008
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?
Solution
The machine completes one third of the job in 
. Thus the entire job is completed in 
. Since the machine is started at 
, the job is finished 
hours later at 
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 | |
