Difference between revisions of "2007 AMC 8 Problems/Problem 1"
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The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so <math>\boxed{\textbf{(D)}\ 12}</math>. | The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so <math>\boxed{\textbf{(D)}\ 12}</math>. | ||
| + | |||
| + | Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
==See Also== | ==See Also== | ||
Revision as of 14:49, 2 July 2024
Contents
Problem
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 
hours per week helping around the house for 
weeks. For the first 
weeks she helps around the house for 
, 
, 
, 
and hours. How many hours must she work for the final week to earn the tickets?
Solution 1
Let 
be the number of hours she must work for the final week. We are looking for the average, so
![\[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\]](http://latex.artofproblemsolving.com/4/5/7/457a1f6e032fdaed73d94270ea967818ce3e6fd8.png)
Solving gives:
![\[\frac{48 + x}{6} = 10\]](http://latex.artofproblemsolving.com/3/0/a/30a8628b16500da9cd381bfecc513d948b887d56.png)
![\[48 + x = 60\]](http://latex.artofproblemsolving.com/4/f/4/4f4e15153e7444b765a0c52ef4902ac9003264e6.png)
![\[x = 12\]](http://latex.artofproblemsolving.com/8/9/4/894d41d4b2af608a93a1b87a8b10fc2097d414ea.png)
So, the answer is 
Solution 2
Use average deviation:
The average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.
So we got
![\[-2+1-3+2+0=-2\]](http://latex.artofproblemsolving.com/9/3/8/93896faef8892a883d0e4a0aeadc1c0896356be6.png)
The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so .
Video Solution by SpreadTheMathLove== https://www.youtube.com/watch?v=omFpSGMWhFc
See Also
| 2007 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. 
