henilshaht wrote:There are 3 different positive integers. If their average (the arithmetic mean) is 8, what are their values?
1) The largest integer is twice the smallest integer.
2) One of them is 9.
Let's take each statement one by one.
Given that the average of 3 different positive integers is 8, the sum of these 3 different positive integers = 3*8 = 24.
1) The largest integer is twice the smallest integer.
Say the smallest integer is x; thus, the largest integer = 2x. Again, say, the middle integer = y. Thus, x < y < 2x.
x + y + 2x = 24
3x + y = 24
Though this is a linear equation with two variables, and it seems that we cannot get unique values of x and y, it is not necessarily so. We must utilize the following two pieces of information.
1. x < y < 2x;
2. x and y are positive integers
Since average = 8, x < 8 such that 2x > 8; thus, 4 < x < 8. Thus, x can be one among 5, 6, and 7.
Case 1: Say x = 5, then from 3x + y = 24, we have y = 9. So the three integers are 5, 9, and 10.
Case 2: Say x = 6, then from 3x + y = 24, we have y = 6 and 2x = 12. This is not possible since we know that 2x > y.
There is no need to check for x = 7 since Case 2 is invalid.
So, only Case 1 is valid; thus, the three integers are 5, 9, and 10. Sufficient.
2) One of them is 9.
We already discussed above that the smallest integer must be less than 8 and the largest integer must be greater than 8.
Case 1: Say the largest integer = 9. Thus, the three integers are 7, 8 and 9.
Case 2: Say the middle integer = 9. Thus, the sum of the smallest and the largest integers is 24 - 9 = 15. The largest integer must be 10 or greater. Possibilities are: {5, 9, 10}; {4, 9, 11}; {3, 9, 12}; {2, 9, 13}; and {1, 9, 14}.
No unique answer. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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