Data Sufficiency

The range of set A is R. A number having a value equal to R, is added to set A. Will the range of set A increase?

(1) All numbers in Set A are positive.
(2) The mean of the new set is smaller than R.

Answer after soem discussion...

Is there an easy systematic approach??

i would say its C

stem 1: insuff

consider 5, 10; if 5 is added then range does not change -- so N
consider 4, 6; if 2 is added then range does increase - so Y

stem 2: insuff

consider 1, 4; if 3 is added then 8/3(new mean)<3 (so condition satisfied) but no change in range -- so N
consider -1, -2; if 1 is added then -2/3(new mean)<1(so condition satisfied) but range increased - so Y

stem 1 and stem 2 together ---- only positive numbers with condition that new mean < R

try 1,4 OR 1,5 --- for instance --- range does not increase if R is added as it will fall between the extremes. I would go with C.

RACHVIK wrote:The range of set A is R. A number having a value equal to R, is added to set A. Will the range of set A increase?

(1) All numbers in Set A are positive.
(2) The mean of the new set is smaller than R.

Answer after soem discussion...

Is there an easy systematic approach??

I received a PM asking me to comment.

Let A = {L,H}. L = lowest value, H = highest value.
R = H-L = value to be added to the set.

If L and H are negative, then the inclusion of R (which is positive, since the range of a set is by definition positive) will increase the range.

If L and H are positive, then in order for the range to increase when R is added, it must be true that R < L.
Thus, for the range to increase:
H-L < L
H < 2L.

So, given two positive numbers in A, the question can be rewritten:

Is H < 2L?

Statement 1: All the numbers in set A are positive.
No way to determine whether H < 2L.
Insufficient.

Statement 2: The mean of the new set is smaller than R.
Let M = mean of new set.
M = (H + L + (H-L))/3 = 2H/3.

If all the numbers are negative, then M = 2H/3 will be negative no matter what values are chosen. As noted above, if L and H are negative, then the inclusion of R (which is positive) will increase the range.

If all the numbers are positive, since M < R, we get:
2H/3 < H-L
2H < 3H - 3L
H > 3L.
If all the numbers are positive and H > 3L, then it is not true that H < 2L, so the inclusion of R will not increase the range.
Insufficient.

Statements 1 and 2 together:
If all the numbers are positive and H > 3L, then it is not true that H < 2L, so the inclusion of R will not increase the range.
Sufficient.

The correct answer is C.

OA is C.

Thanks everyone for your reply.