Difference between revisions of "2015 AMC 10B Problems/Problem 2"

(See Also)
(Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
 
Marie does her work twice 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)}3:30PM}</math>
 
Marie does her work twice 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)}3:30PM}</math>
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=B|before=Problem 1|num-a=3}}
 +
{{MAA Notice}}

Revision as of 18:42, 3 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

Marie does her work twice 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 $\boxed{\textbf{(B)}3:30PM}$

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

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

Invalid username
Login to AoPS