This one took me a moment . Start with this idea: if the process works for any number of numbers in the set, then we might as well work it out for ONLY ONE NUMBER. In other words, n = 1.

If we have one number with the sum S, then we adjust as follows:

S + 3

then

(S + 3) * 5

then

5(S+3) / 7

This gives us:

(5S + 15) / 7

or

5S/7 + 15/7, and we are told that these must be greater than the sum of the original set, or S.

That is, 5S/7 + 15/7 > S

Put all the S-terms together:

15/7 > 2S/7

Multiply this by 7:

15 > 2S, or 2S < 15. Remember that n = 1, so we can say 2S < 15n.

Answer B.