cramya wrote:
Stmt I

x^2 + 13x = 7 mean

x(x+13) = 7 * mean

x can be 1 or a multiple of 7 and the mean changes

Brent Hanneson wrote:I need to sign off, so were's my solution:
The mean of set A is (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2)/7 = (x^2+13x)/7 = x(x+13)/7
(1) If the mean is less than 15 then x(x+13)/7 < 15, which means that x(x+13) < 105
We can see that, if x(x+13) < 105 then x < 6 (i.e, x = 1, 2, 3, 4, 5 or 6)
Only one of the these values (x=1) yields a integer mean
So, x must equal 1 (sufficient)
(2) If we add 10-16x to set A we get: mean = (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2 + 10-16x)/8
Simplify to get: (x^2 - 3x + 10)/8 = 1 --> x^2 - 3x + 10 = 8 --> x^2 - 3x + 2 = 0 --> x=1 or 2 (insuff)

The answer is A