#Here, The average of 7 numbers is 12 & The average of the 4 smallest numbers in this set is 8, while the average of the 4 greatest numbers in this set is 20.That means 1 number should be overlapped.
Now,
(A+B+C+D)/4=8, A+B+C+D=32..........(1)
(D+E+F+G)/4=20; D+E+F+G=80........(2)
#(2)-(1).=
D+(E+F+G)-(A+B+C)-D=80-32=48 Answer(D)

Looks like there is an issue with the question. Assuming the numbers in ascending order are A,B,C,D,E,F,G. Assuming D<E<F<G. How can D, the smallest among the 4 greatest numbers be equal to 28 while the average of the 4 greatest numbers equals 20? Average cannot be smaller than the smallest number of the group right?

A+B+C+D+E+F+G=7*12=84
(A+B+C+D)+(D+E+F+G)=4*8+4*20=112

Subtracting the first from second, we get D=112-84=28

if D is 28, how can the average of D and three larger numbers be equal 20? hence I think the question is wrongly framed and inconsistent with the solution.

multiple edits because my keyboard was acting weird during edits.

Correct

But the question is written correctly

I dont see any issue with the question, but first where did you see that the value of d is 28?

refer to the solution above, the fourth number in the series should be 28 based on the totals and the averages given

Oh...I get your point now. The algebra may be correct in getting the answer but the question itself appears to be mathematically disjointed. Carcass, what have you got to say to this sir?

If $\(x_4\)$ is the smallest of the 4 greatest numbers, then $\(x_5, x_6\)$, and $\(x_7\)$ must all be greater than or equal to $\(x_4\)$.
If $\(x_4=28\)$, then the minimum possible sum for the 3 greatest numbers is:

$$
\(3 \times 28=84\)
$$


But the problem requires their sum to be $\(\mathbf{5 2}\)$. Since $\(52<84\)$, it is impossible to construct this list. The average of the " 4 greatest" is mathematically incompatible with the total average given the required ordering of the numbers.

I think this