Difference between revisions of "2011 AMC 8 Problems/Problem 5"
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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>. | 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) | MegaBoy6679 :D 23:31, 26 December 2022 (EST) | ||
Revision as of 23:31, 26 December 2022
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)
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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