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 <math>2</math> 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> |
==Video Solution== | ==Video Solution== | ||
Revision as of 16:04, 13 July 2020
Contents
Problem
Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 
PM and finishes the second task at 
PM. When does she finish the third task?
Solution
Marie finishes 
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. 
