Difference between revisions of "2011 AMC 8 Problems/Problem 5"
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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>. | 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>. | ||
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-RealCXY | -RealCXY | ||
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| + | ==Solution 3 (Estimate)== | ||
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| + | The answer choices are relatively far apart, so we can use estimation to solve this problem. <math>2100/60</math> is around <math>35</math> hours, or about <math>1.5</math> days. The only answer choice around <math>1.5</math> days from the midnight of January 1, 2011 is <math>\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}</math>. | ||
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| + | MegaBoy6679 :D 23:31, 26 December 2022 (EST) | ||
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/G4TaNyvDUOQ | ||
==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}} | ||
Latest revision as of 10:03, 18 November 2024
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
Solution 3 (Estimate)
The answer choices are relatively far apart, so we can use estimation to solve this problem. 
is around 
hours, or about 
days. The only answer choice around days from the midnight of January 1, 2011 is 
.
MegaBoy6679 :D 23:31, 26 December 2022 (EST)
Video Solution by WhyMath
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. 
