Difference between revisions of "1996 AJHSME Problems/Problem 12"
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| − | Adding all of the numbers gives us | + | ==Problem 12== |
| + | |||
| + | What number should be removed from the list | ||
| + | <cmath>1,2,3,4,5,6,7,8,9,10,11</cmath> | ||
| + | so that the average of the remaining numbers is <math>6.1</math>? | ||
| + | |||
| + | <math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math> | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <math>\frac{66-x}{10}=6.1</math> | ||
| + | |||
| + | <math>66 - x = 61</math> | ||
| + | |||
| + | <math>x = 5</math> | ||
| + | |||
| + | Thus, the answer is <math>\boxed{B}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1996|num-b=11|num-a=13}} | ||
| + | * [[AJHSME]] | ||
| + | * [[AJHSME Problems and Solutions]] | ||
| + | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:24, 4 July 2013
Contents
Problem 12
What number should be removed from the list
![\[1,2,3,4,5,6,7,8,9,10,11\]](http://latex.artofproblemsolving.com/3/9/9/3997f2316718e3ba0050cf5098486f6f7799080d.png)
so that the average of the remaining numbers is 
?
Solution 1
Adding all of the numbers gives us 
as the current total. Since there are 
numbers, the current average is 
. We need to take away a number from the total and then divide the result by 
because there will only be numbers left to give an average of . Setting up the equation:
Thus, the answer is 
Solution 2
Similar to the first solution, the current total is 
. Since there are numbers on the list, taking 
number away will leave numbers. If those numbers have an average of , then those numbers must have a sum of 
. Thus, the number that was removed must be 
, and the answer is .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
