Difference between revisions of "2016 AMC 8 Problems/Problem 1"

m (Solution)
m
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes.  We know that there is <math>60</math> minutes in a hour.  Therefore, there are <math>11 \cdot 60 = 600 + 60 = 660</math> minutes in 11 hours.  Adding the second part(the 5 minutes) we get <math>660 + 5 = \boxed{665}</math>.
+
It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes.  We know that there is <math>60</math> minutes in a hour.  Therefore, there are <math>11 \cdot 60 = 660</math> minutes in 11 hours.  Adding the second part(the 5 minutes) we get <math>660 + 5 = \boxed{665}</math>.
  
 
{{AMC8 box|year=2016|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2016|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:38, 27 November 2016

The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?

$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$

Solution

It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{665}$.

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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 AJHSME/AMC 8 Problems and Solutions

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