Difference between revisions of "2016 AMC 8 Problems/Problem 1"
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==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 = 660</math> minutes in 11 hours. Adding the second part(the 5 minutes) we get <math>660 + 5 = \boxed{665}</math>. | + | 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 = 660</math> minutes in 11 hours. Adding the second part(the 5 minutes) we get <math>660 + 5 = \boxed{\textbf{(C)}\ 665}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of <math>\boxed{{(C) } | ||
| + | 665}</math>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | 11 hours 5 min = (11 * 60) + 5 min = 665 min, therefore | ||
| + | <math>\boxed{{(C) } | ||
| + | }</math>. | ||
| + | |||
| + | Carmen | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
| + | |||
{{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 00:44, 7 November 2020
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(the 5 minutes) we get 
.
Solution 2
The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of 
.
Solution 3
11 hours 5 min = (11 * 60) + 5 min = 665 min, therefore
.
Carmen
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
| 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. 
