Difference between revisions of "2015 AMC 12B Problems/Problem 2"

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==Problem==
 
==Problem==
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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?
  
 
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<math> \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}</math>
  
 
==Solution==
 
==Solution==
 
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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</math> minutes <math>=\fbox{\textbf{(B)}\; \text{3:30 PM}}</math>.
  
 
==See Also==
 
==See Also==
{{AMC12 box|year=2015|ab=B|before=1|num-a=3}}
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{{AMC12 box|year=2015|ab=B|num-a=3|num-b=1}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:05, 8 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?

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

Solution

The first two tasks took $\text{2:40 PM}-\text{1:00 PM}=100$ minutes. Thus, each task takes $100\div 2=50$ minutes. So the third task finishes at $\text{2:40 PM}+50$ minutes $=\fbox{\textbf{(B)}\; \text{3:30 PM}}$.

See Also

2015 AMC 12B (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 12 Problems and Solutions

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