Difference between revisions of "2011 AMC 8 Problems/Problem 5"
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<math>\textbf{(E)}\ \text{January 2 at 6:01PM}</math> | <math>\textbf{(E)}\ \text{January 2 at 6:01PM}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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> | ||
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| + | ==Solution 2== | ||
| + | |||
| + | Like the previous solution, <math>2011</math> minutes is <math>33</math> hours and <math>31</math> minutes. That means it has to be January 1 at 9:31PM or January 2 at 9:31AM because they are the only ones that end with 31. But, the answer can’t be in January 1 because there are 24 hours in a day and 33 is greater than 24. So, the answer is <math>\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}</math>. | ||
| + | -RealCXY | ||
==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 22:09, 26 October 2021
Contents
Problem
What time was it 
minutes after midnight on January 1, 2011?
Solution 1
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. 
Solution 2
Like the previous solution, minutes is hours and minutes. That means it has to be January 1 at 9:31PM or January 2 at 9:31AM because they are the only ones that end with 31. But, the answer can’t be in January 1 because there are 24 hours in a day and 33 is greater than 24. So, the answer is .
-RealCXY
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
