A series of 5 numbers is 3, 4, 5, 5, x, is the range greater than 2?
1. the median of the numbers is greater than the mean
2. the median is 4

Sum of the 5 numbers = x+3+4+5+5 = x+17.
In each case, test whether it's possible for the range of the 5 numbers to be EQUAL TO 2.

Statement 1: the median of the numbers is greater than the mean
Case 1: median = 4
If x=3, then the range is equal to 2 and the set looks as follows: x=3, 3, 4, 5, 5.
Average = (3+17)/5 = 4.
Not viable: the average must be LESS than the median.
To decrease the average, x must be LESS than 3.
Thus, the set must look as follows:
x<3, 3, 4, 5, 5.
Since the smallest value is less than 3 and the greatest value is 5, the range of the 5 numbers is GREATER THAN 2.

Case 2: median = 5
If x=5, then the range is equal to 2 and the set looks as follows: 3, 4, 5, 5, x=5.
Average = (5+17)/5 = 22/5 = 4.4.
This works, since the average is less than the median.
Since the smallest value is 3 and the greatest value is 5, the range of the 5 numbers is EQUAL TO 2.

Since the range is greater than 2 in Case 1 but equal to 2 in Case 2, INSUFFICIENT.

Statement 2: the median is 4
Case 1 also satisfies statement 2.
In Case 1, the range > 2.

Case 3:
If x=5, then the range is EQUAL TO 2 and the set looks as follows: x=3, 3, 4, 5, 5.

Since the range is greater than 2 in Case 1 but equal to 2 in Case 3, INSUFFICIENT.

Statements combined:
Both statements are satisfied only by Case 1.
Thus, the range > 2.
SUFFICIENT.

The correct answer is C.

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A series of 5 numbers is 3, 4, 5, 5, x, is the range greater than 2?
1> the median of the numbers is greater than the mean
2> the median is 4

There is one variable (x) so we need one equation in order to solve for the variable, but 2 equations are given by the conditions, so there is high chance that (D) is going to be our answer.
For condition 1: 4,x,5 is possible for median, and the average becomes (3+4+5+5+x)/5=(17+x)/5, 4>(17+x)/5, 3>x, so the range is always greater than 2. The question can be answered 'yes' and from x>(17+x)/5, we can obtain 4x>17, x>4.25, 4.25<x<=5, so the question is answered 'no'. In other words, the answer to the question is 'no' for x=5. Hence, the conditions are insufficient.
For condition 2, in order to let 4 be the median, range=5-3=2>2 which becomes 'no', but 'yes' for x=1, range=5-1=4>2. This condition is, again, insufficient.
Looking at the conditions together, x<=4 and for x<3, the range is always greater than 2, which answers the question 'yes' therefore the combined condition is sufficient, and the answer becomes (C).

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.

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C