Difference between revisions of "2016 AMC 8 Problems/Problem 1"
Jacoporizzo (talk | contribs) (→Solution 2) |
Elijahman112 (talk | contribs) (→Video Solution (THINKING CREATIVELY!!!)) |
||
| (12 intermediate revisions by 7 users not shown) | |||
| Line 1: | Line 1: | ||
| + | == Problem == | ||
The longest professional tennis match ever played lasted a total of <math>11</math> hours and <math>5</math> minutes. How many minutes was this? | The longest professional tennis match ever played lasted a total of <math>11</math> hours and <math>5</math> minutes. How many minutes was this? | ||
<math>\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105</math> | <math>\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105</math> | ||
| − | ==Solution== | + | == 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 <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>. | 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== | + | ===Solution 2=== |
| − | The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of <math>\boxed{{(C)} 665}</math>. | + | 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 = <math>(11 \cdot 60) + 5 \text{min} = 665 \text{min}</math>, therefore | ||
| + | <math>\boxed{{(C) } | ||
| + | }</math>. | ||
| + | |||
| + | 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== | ||
| + | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
| + | |||
| + | https://youtu.be/1cu8fT2UJCw | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==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. 
