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Difference between revisions of "2007 AMC 8 Problems/Problem 1"

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Let <math>x</math> be the number of hours she must work for the final week. We are looking for the average, so  <math>\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10</math>   
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Let <math>x</math> be the number of hours she must work for the final week. We are looking for the average, so   
−
 
+
<cmath>\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10</cmath>   
 
Solving gives:
 
Solving gives:
−
 
+
<cmath>\frac{48 + x}{6} = 10</cmath>  
−
<math>\frac{48 + x}{6} = 10</math>  
 
 
   
 
   
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<math>48 + x = 60</math>
+
<cmath>48 + x = 60</cmath>
  
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<math>x = 12</math>
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<cmath>x = 12</cmath>
  
 
So, the answer is <math> \boxed{\textbf{(D)}\ 12} </math>
 
So, the answer is <math> \boxed{\textbf{(D)}\ 12} </math>

Revision as of 07:20, 27 August 2015

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

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

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

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