Difference between revisions of "2011 AMC 8 Problems/Problem 5"

Line 13: Line 13:
 
==Solution==
 
==Solution==
  
There are <math>60</math> minutes in an hour. <math>2011/60=33\text{r}31,</math> or <math>33</math> hours and <math>31</math> minutes. There are <math>24</math> hours in a day, so the time is <math>9</math> hours and <math>31</math> minutes after midnight on January 2, 2011. <math>\Rightarrow \boxed{\textbf{(D)}}</math>
+
There are <math>60</math> minutes in an hour. <math>2011/60=33\text{r}31,</math> or <math>33</math> hours and <math>31</math> minutes. There are <math>24</math> hours in a day, so the time is <math>9</math> hours and <math>31</math> minutes after midnight on January 2, 2011. <math>\Rightarrow \boxed{\textbf{(D)}\ \text{January 2 at 9:31AM} }}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=4|num-a=6}}
 
{{AMC8 box|year=2011|num-b=4|num-a=6}}

Revision as of 18:38, 4 November 2012

What time was it $2011$ minutes after midnight on January 1, 2011?

$\textbf{(A)}\ \text{January 1 at 9:31PM}$

$\textbf{(B)}\ \text{January 1 at 11:51PM}$

$\textbf{(C)}\ \text{January 2 at 3:11AM}$

$\textbf{(D)}\ \text{January 2 at 9:31AM}$

$\textbf{(E)}\ \text{January 2 at 6:01PM}$

Solution

There are $60$ minutes in an hour. $2011/60=33\text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $\Rightarrow \boxed{\textbf{(D)}\ \text{January 2 at 9:31AM} }}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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
Invalid username
Login to AoPS