Difference between revisions of "2011 AMC 8 Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | Average the differences between each day. We get <math>10, -10,\text{ } 20,\text{ } 30,-20</math>. We find the average of this list to get <math>\boxed{\textbf{(A)}\ 6}</math>. | + | Average the differences between each day. We get <math>10, -10,\text{ } 20,\text{ } 30,-20</math>. We find the average of this list to get <math>\boxed{\textbf{(A)}\ 6}</math>. ( In case you were wondering, the way to calculate the average is <math>\frac{(10+(-10)+20+30+(-20))}{5} = \frac{ 30}{5} = 6</math>. So the answer is indeed <math>\boxed{\textbf{(A)}\ 6}</math>) |
==Solution 2== | ==Solution 2== | ||
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-MiracleMaths | -MiracleMaths | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=10|num-a=12}} | {{AMC8 box|year=2011|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:00, 17 December 2023
Problem
The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
Solution
Average the differences between each day. We get . We find the average of this list to get
. ( In case you were wondering, the way to calculate the average is
. So the answer is indeed
)
Solution 2
This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day). 86 - 80 = 6. Therefore, the answer is .
-MiracleMaths
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.