I assume you mean standard deviation is d, since you don't mention what d is, only that it's positive.

In that case, the answer is e, because standard deviation is a measure of variability, and by adding points in the middle you're decreasing that variability.

All the other answers don't make sense because you could have a set of 50 5's and 50 7's, in which case adding a -6, 0, or 12 would increase the standard deviation.

assuming the SD to be 'd', as its not mentioned.
I think the answer is E as that would decrease the variability of the data.
Also going by the formula that
SD=sqrRoot[sum(X-M)/N], where N is the number of terms, X being the value of each term and M the mean.

If we introduce two elements 6 and 6, the sum(X-M) will not increase as the mean itself is 6, but N has increased by 2 so the overall SD would decrease.


Cheers,
Vivek

d^2=[(a1-a)^2+(a2-a)^2+....+(an-a)^2 ]/n
When we added 6 and 6, the numerator remained unchanged but the denominator
increased, so, the new deviation is less than d.


other way ( through negation technique)

choose the option one by one, assume the mean=average=6
1) -6, and 0 , since the sd is distance from the mean
-6 and 0 will increase the distance from the mean hence the data will more spreaded across the mean and hence sd will increase. ( so negate the option)
2)0 and 0 , since the sd is the distance from the mean .Here the SD will increase .
3)0 and 6 , the sd will increase, since sd is the distance from the mean
4)0 and 12, the sd will increase , since sd is the distance from the mean
5) 6, and 6 ,the sd will remain as it is; which means negligible data spread from the mean 6.


most of the time SD characteristics help to solve the question very quickly which might take more minutes when solving in any other way.

thanks,
surbhi