Difference between revisions of "2016 AMC 8 Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
− | It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is <math>60</math> minutes in a hour. Therefore, there are <math>11 \cdot 60 = 600 + 60 = 660</math> minutes in 11 hours. Adding the second part, we get <math>660 + 5 = | + | It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is <math>60</math> minutes in a hour. Therefore, there are <math>11 \cdot 60 = 600 + 60 = 660</math> minutes in 11 hours. Adding the second part, we get <math>660 + 5 = 665</math> or (). |
{{AMC8 box|year=2016|before=First Problem|num-a=2}} | {{AMC8 box|year=2016|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:47, 23 November 2016
The longest professional tennis match ever played lasted a total of hours and minutes. How many minutes was this?
Solution
It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is minutes in a hour. Therefore, there are minutes in 11 hours. Adding the second part, we get or ().
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.