# 1996 AJHSME Problems/Problem 12

## 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 $\boxed{B}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 