Difference between revisions of "2015 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
Marie finishes 2 tasks 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 finishes 2 tasks 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> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/7CcxOXjv3i8 | ||
| + | |||
| + | ~savannahsolver | ||
==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 17:04, 16 June 2020
Contents
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 finishes 2 tasks in 
hour and 
minutes. Therefore, one task should take 
minutes to finish. minutes after 
PM is 
PM, so our answer is 
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
