2015 AMC 12B Problems/Problem 2

Revision as of 09:02, 4 March 2015 by Pi over two (talk | contribs) (Solution)

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 $\text{2:40 PM}+50=\fbox{\textbf{(B)}\; \text{3:30 PM}}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png