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Difference between revisions of "1996 AJHSME Problems/Problem 12"

(Created page with "Adding all of the numbers gives us (11)(12/2)=66. Since there are 11 numbers, the average is 66/11=6. We need to take away a number from the total sum and then divide the result ...")
 
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Adding all of the numbers gives us (11)(12/2)=66. Since there are 11 numbers, the average is 66/11=6. We need to take away a number from the total sum and then divide the result by 10 because there will only be 10 numbers left to give us 6.1. Setting up the equation, we get (66-x)/10=6.1. Solving for x gives 5 which is choice B.
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==Problem 12==
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What number should be removed from the list
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<cmath>1,2,3,4,5,6,7,8,9,10,11</cmath>
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so that the average of the remaining numbers is <math>6.1</math>?
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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
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==Solution 1==
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Adding all of the numbers gives us <math>\frac{11\cdot12}{2}=66</math> as the current total. Since there are <math>11</math> numbers, the current average is <math>\frac{66}{11}=6</math>. We need to take away a number from the total and then divide the result by <math>10</math> because there will only be <math>10</math> numbers left to give an average of <math>6.1</math>. Setting up the equation:
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<math>\frac{66-x}{10}=6.1</math>
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<math>66 - x = 61</math>
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<math>x = 5</math>
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Thus, the answer is <math>\boxed{B}</math>
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==Solution 2==
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Similar to the first solution, the current total is <math>66</math>.  Since there are <math>11</math> numbers on the list, taking <math>1</math> number away will leave <math>10</math> numbers.  If those <math>10</math> numbers have an average of <math>6.1</math>, then those <math>10</math> numbers must have a sum of <math>10 \times 6.1 = 61</math>. Thus, the number that was removed must be <math>66 - 61 = 5</math>, and the answer is <math>\boxed{B}</math>.
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==See Also==
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{{AJHSME box|year=1996|num-b=11|num-a=13}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]

Revision as of 16:46, 1 August 2011

Problem 12

What number should be removed from the list \[1,2,3,4,5,6,7,8,9,10,11\] so that the average of the remaining numbers is $6.1$?

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution 1

Adding all of the numbers gives us $\frac{11\cdot12}{2}=66$ as the current total. Since there are $11$ numbers, the current average is $\frac{66}{11}=6$. We need to take away a number from the total and then divide the result by $10$ because there will only be $10$ numbers left to give an average of $6.1$. Setting up the equation:

$\frac{66-x}{10}=6.1$

$66 - x = 61$

$x = 5$

Thus, the answer is $\boxed{B}$

Solution 2

Similar to the first solution, the current total is $66$. Since there are $11$ numbers on the list, taking $1$ number away will leave $10$ numbers. If those $10$ numbers have an average of $6.1$, then those $10$ numbers must have a sum of $10 \times 6.1 = 61$. Thus, the number that was removed must be $66 - 61 = 5$, and the answer is $\boxed{B}$.

See Also

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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