Hi,
Let the mean be x. If all four number are multiplied by a constant say k, the mean will be kx.
But, given that kx = x for all k.
So, (k-1)x= 0for all k.
So, x is zero.
So, I is true
II)The set is -3,-2,-1,6 will satisfy this. But, the sum of largest and smallest is not zero
False
III)The set of 0,0,0,0 also satisfies
False

Hence, A

Mean* Constant = Mean So, Mean has to be zero

To have mean zero, we have to have both negative and positive numbers

The sum of sallest and biggest does not have to be zero. -4 -1 2 3 or -2 -1 1 2

Ans D in my opinion.

Oh, I missed that 0000 set. Since it does not say different integers, that is right.

a necessary and sufficient kind of question this is.

necessary condition is that the mean has to be 0.
sufficient conditions are the sum should be zero for the numbers and the set may/may not have positive and negative numbers.

tip is to pick up the necessary condition here. Hence A.

Below is my approach
Could anyone please tell me if I missed something or something inappropriate.
(x1 + x2 + x3 +x4)/4 = K
-->x1 + x2 + x3 +x4 = 4K
Let a be the constant number
a*x1 + a*x2 + a*x3 + a*x4 = 4K

(1) The mean of the set is 0, which means the sum is 0
x1 + x2 + x3 + x4 = 0 and
a*x1 + a*x2 + a*x3 + a*x4 = 0
x1 + x2 + x3 + x4 = a*x1 + a*x2 + a*x3 + a*x4
x1 + x2 + x3 + x4 = a*( x1 + x2 + x3 + x4)--> equ(1)
equation (1) gives us 1 possible outcome:
x1 + x2 + x3 + x4 = 0
Hence, (1) is always true

(2) The sum of the largest member and the smallest member of the set is 0
Take some examples: -8 + 4 + 0 + 8 = 4
And -4 + 2 + 0 + 4 = 2
The sum varies due to the number we choose-->(2) is not true

(3) The set contains both positive and negative integers
From examples above, the sets contain both positive number and negative number, but their sums do not equal a constant number. Hence, (3) is not true.
Choose A

Thank you!!