We have two sets, with the terms ordered from the least to the largest:

$$A=a_1,a_2,a_3,a_4,a_5$$

And

$$B=b_1,a_2,a_3,a_4,a_5$$

And

$$b_1>a_1$$

Then the median ($a_3$) will be same.

Since $b_1>a_1$, the range of B ($a_5-b_1$) < the range of A ($a_5-a_1$).

Since $b_1>a_1$,

$$MeanofB=\frac{b_1+a_2+a_3+a_4+a_5}{5}$$

$$MeanofA=\frac{a_1+a_2+a_3+a_4+a_5}{5}$$

$$MeanofA-MeanofB=\frac{a_1+a_2+a_3+a_4+a_5-(b_1+a_2+a_3+a_4+a_5)}{5}$$

$$=\frac{a_1-b_1}{5}<0$$

$$=>MeanofA-MeanofB<0$$

$$\therefore MeanofA<MeanofB$$

Only the mean must be greater for set B than for set A. Therefore A is correct.