Difference between revisions of "2011 AMC 8 Problems/Problem 5"
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==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)} | + | 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 minutes after midnight on January 1, 2011?
Solution
There are minutes in an hour. or hours and minutes. There are hours in a day, so the time is hours and minutes after midnight on January 2, 2011.
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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