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, 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>
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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>
  
 
==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 15:40, 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?

$\textbf{(A) }\text{3:10 PM}\qquad\textbf{(B) }\text{3:30 PM}\qquad\textbf{(C) }\text{4:00 PM}\qquad\textbf{(D) }\text{4:10 PM}\qquad\textbf{(E) }\text{4:30 PM}$

Solution

Marie finishes 2 tasks in $1$ hour and $40$ minutes. Therefore, one task should take $50$ minutes to finish. $50$ minutes after $2\!:\!40$ PM is $3\!:\!30$ PM, so our answer is $\boxed{\textbf{(B) }\text{3:30 PM}}$

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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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