Difference between revisions of "2015 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | Marie does her work twice in <math>1</math> hour and <math>40</math> minutes. Therefore, one task should take <math>50</math> minutes to finish. <math>50</math> minutes after <math>2\!:\!40</math> PM is <math>3\!:\!30</math> PM, so our answer is <math>\boxed{\textbf{(B) }\text{3:30 PM}}</math> | + | Marie does her work twice in <math>1</math> hour and <math>40</math> minutes, or 100 minutes. Therefore, one task should take <math>50</math> minutes to finish. <math>50</math> minutes after <math>2\!:\!40</math> PM is <math>3\!:\!30</math> PM, so our answer is <math>\boxed{\textbf{(B) }\text{3:30 PM}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2015|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:36, 10 February 2019
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
Marie does her work twice in 
hour and 
minutes, or 100 minutes. Therefore, one task should take 
minutes to finish. minutes after 
PM is 
PM, so our answer is 
See Also
| 2015 AMC 10B (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 10 Problems and Solutions | ||
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