If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,

BTGmoderatorDC wrote:If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?

A. 999
B. 1001
C. 1003
D. 1004
E. 1005

Sum of the 10 terms = (quantity)(average) = 10*999 = 9990.
If we add the units digits of the nine given terms, we get:
3+4+6+7+8+1+1+2+4 = 36.
The sum of the nine given terms has a units digit of 6.
The sum of all ten terms has a units digit of 0.
Since (units digit of 6) + (units digit of 4) = (units digit of 0), x must have a units digit of 4.

The correct answer is D.

BTGmoderatorDC wrote:If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?

A. 999
B. 1001
C. 1003
D. 1004
E. 1005

Since the numbers are “pretty” large and each is “pretty” close to the average, without using the conventional formula (i.e., average = sum/quantity), we can determine the value of x by using the difference between each known number and the average. If we subtract 999, the average, from each number, we have:

-6, -5, -3, -2, -1, 2, 2, 3, 5

We see that the sum of the differences is (-17) + 12 = -5. To counter the -5 (i.e., 5 less than the average), we need a value that is 5 more than the average. That is, we need x to be 999 + 5 = 1004 so that the average of all values (including x) will be 999.

Answer: D