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 $\boxed{B}$ .

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
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1996 AJHSME Problems/Problem 12

Contents

Problem 12

Solution 1

Solution 2

See Also