2007 AMC 8 Problems/Problem 1

Problem

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ , $11$ , $7$ , $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?

$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$

Solution 1

Let $x$ be the number of hours she must work for the final week. We are looking for the average, so $\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10$ Solving gives: $\frac{48 + x}{6} = 10$

$48 + x = 60$

$x = 12$

So, the answer is $\boxed{\textbf{(D)}\ 12}$

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

$-2+1-3+2+0=-2$

The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so $\boxed{\textbf{(D)}\ 12}$ .

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by WhyMath

https://youtu.be/Y6LrJn1Kuvg

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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