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 $\boxed{\textbf{(D)}\ 12}$.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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

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