Difference between revisions of "2015 AMC 12B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | The first two tasks took <math>2:40-1:00=100</math> minutes. Thus, each task takes <math>100\div 2=50</math> minutes. So the third task finishes at <math>2:40+50=3:30</math>. | + | The first two tasks took <math>\text{2:40 PM}-\text{1:00 PM}=100</math> minutes. Thus, each task takes <math>100\div 2=50</math> minutes. So the third task finishes at <math>\text{2:40 PM}+50=\fbox{\textbf{(B)}\; \text{3:30 PM}}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=3|num-b=1}} | {{AMC12 box|year=2015|ab=B|num-a=3|num-b=1}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:02, 4 March 2015
Problem
Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?
Solution
The first two tasks took 
minutes. Thus, each task takes 
minutes. So the third task finishes at 
.
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
