Be careful with data sufficiency! If we come up with a definite answer to the question given a statement, then the statement is sufficient. Therefore, we should try to arrive at multiple conclusions to prove the statement insufficient. For a Yes/No question, this means finding a YES case and a NO case. For value questions, this means finding multiple values that could answer the question.
Indeed, the range of a set is the difference between the largest value and the smallest. For example, if our set is {5, 21, 25}, our range would be 25-5=20.
Statement 1: All numbers in Set A are positive.
NO case: {2, 7, 10}
R = 10-2 = 8
Our new set is {2, 7, 8, 10}
Range of new set = 10-2 = 8
Therefore, our range does not change.
YES case: {22, 27, 30}
R = 30-22 = 8
Our new set is {8, 22, 27, 30}
Range of new set = 30-8 = 22
Therefore, the range of our set INCREASES.
Because we have a YES case and a NO case, statement 1 is INSUFFICIENT. In other words, we can not answer definitively YES or definitively NO.
Statement 2: The mean of the new set is smaller than R.
Here are three examples of sets that have means smaller than their respective ranges:
Example Set 1: {1,3,5,7}
Mean = 4, R = 6
Example Set 2: {0, 10, 12, 18}
Mean = 10, R = 18
Example Set 3: {-10, -8, 2, 4}
Mean = -3, R = 14
Again, our task is to prove the statement insufficient by finding a YES case and a NO case. We think to ourselves, "Given that mean<range in the new set, can I find a case in which the range increases? Can I find a case in which the range does not increase?"
NO case: Take Example Set 1 from above: {1,3,5,7}
Mean = 4, R = 6
Our new set is {1, 3, 5, 6, 7}
Range = 6
Mean of new set = (1+3+5+6+7)/5 = 22/5 = 4.4
4.4<6
Our range does not increase.
YES case: Take Example Set 3 from above: {-10, -8, 2, 4}
Mean = -3, R = 14
Our new set is {-10, -8, 2, 4, 14}
Range of new set = 14-(-10) = 24
Mean of new set = 2/5 = 0.4
0.4<14
Our range does increase.
Because we have a YES case and a NO case, statement 2 is INSUFFICIENT. In other words, we can not answer definitively YES or definitively NO.
Both statements together: All #'s in the set are positive, and the mean of the new set is less than R.
NO case: We can again use Example Set 1 from above: {1,3,5,7}
Mean = 4, R = 6
Our new set is {1, 3, 5, 6, 7}
Range of new set = 6 = R
Mean of new set = (1+3+5+6+7)/5 = 22/5 = 4.4
4.4<6
Our range does not increase.
If the mean is less than the range, then we would need to find an R greater than the largest value in the set in order to increase the range. The only way to have an R that is greater than the greatest value in the set is for the set to include at least one negative number. (Note that the range must be positive, so finding an R lower than 0 would be impossible.) Therefore, we cannot find a YES case given Statement 1 and 2 combined, and we have a definitive NO answer. The statements combined are sufficient.
Answer: C
There are probably more intuitive ways, but you can use this explanation as an exercise to build your intuition. Good luck!
