2007 AMC 8 Problems/Problem 1: Difference between revisions

Latest revision as of 16:28, 28 October 2024

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
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

@@ Line 4: / Line 4: @@
 <math>\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13</math>
-== Solution ==
+== Solution 1 ==
-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>
+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:
+<cmath>\frac{48 + x}{6} = 10</cmath>
+<cmath>48 + x = 60</cmath>
+<cmath>x = 12</cmath>
+So, the answer is <math> \boxed{\textbf{(D)}\ 12} </math>
+== 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
+<cmath>-2+1-3+2+0=-2</cmath>
+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
+==Video Solution by WhyMath==
+https://youtu.be/Y6LrJn1Kuvg
+==See Also==
+{{AMC8 box|year=2007|before=First Problem|num-a=2}}
-<math>\frac{48 + x}{6} = 10</math>
+[[Category:Introductory Algebra Problems]]
-<math>48 + x = 60</math>
+{{MAA Notice}}
-<math>x = 12</math>
-So, <math>\boxed{D}</math>

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