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Difference between revisions of "2011 AMC 8 Problems/Problem 5"
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==Solution==
 
==Solution==
  
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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>
+
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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:09, 25 March 2015

Problem

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

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
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

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