Difference between revisions of "2016 AMC 8 Problems/Problem 1"
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===Solution 3=== | ===Solution 3=== | ||
− | 11 hours 5 min = (11 | + | 11 hours 5 min = <math>(11 \cdot 60) + 5 \text{min} = 665 \text{min}</math>, therefore |
<math>\boxed{{(C) } | <math>\boxed{{(C) } | ||
}</math>. | }</math>. | ||
Carmen | Carmen | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/je42D7WIJWw?si=XD3Mf-vaxgyF726G | ||
+ | |||
+ | It's so simple that a 12-year-old Made the Video! | ||
+ | |||
+ | ~Elijahman~ | ||
+ | |||
+ | ==Video Solution (THINKING CREATIVELY!!!)== | ||
+ | https://youtu.be/-T0AZIsu4DE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
+ | https://youtu.be/1cu8fT2UJCw | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|before=First Problem|num-a=2}} | {{AMC8 box|year=2016|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:35, 20 June 2024
Contents
Problem
The longest professional tennis match ever played lasted a total of hours and minutes. How many minutes was this?
Solutions
Solution 1
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 = , therefore .
Carmen
Video Solution
https://youtu.be/je42D7WIJWw?si=XD3Mf-vaxgyF726G
It's so simple that a 12-year-old Made the Video!
~Elijahman~
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
~savannahsolver
See Also
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.