DS problem

aspire750 wrote:

nik08 wrote:A data sufficiency question :

Set S contains more than one element. Is the range of the set S larger than its mean?
1) Set S does not conatin positive elements
2) the median of set S is negative

Can someone please explain ?

Thanks in adv!

2) the median of set S is negative- if median of a set is known, the mean of the set equals median. Sufficient

No, the median does not normally equal the mean. The median and mean are sometimes equal- if you have an evenly spaced set, for example, or a symmetrically distributed set- but if you don't know anything about the set, you certainly can't conclude anything about the relationship between the mean and the median. The median of the following set is 1:

{0, 1, 10000000000000000}

but the mean is much larger than 1.

The answer should be C, although A is very, very close to being sufficient.
If we knew that all the elements were negative, we would know that the mean was negative. Since the range of a set can't be negative, if a set only contains negative elements, the range must be larger than the mean. But, that's not quite what Statement 1 says. From 1, it's certainly possible that every element in the set is equal to zero. Then the range is zero and the mean is zero; the mean doesn't need to be smaller than the range.

From 1+2 together, we know the set contains at least one negative element, and does not contain any positive elements, so it must have a negative mean, and therefore the range is larger than the mean.

nik08 wrote:A data sufficiency question :

Set S contains more than one element. Is the range of the set S larger than its mean?
1) Set S does not conatin positive elements
2) the median of set S is negative

Can someone please explain ?

Thanks in adv!

My thanks to shalen78, who pointed out via pm that I didn't even consider the possibility that 2) is sufficient alone- a classic DS mistake! And 2) is sufficient, alone, in fact. The correct answer is B. If we know the median is negative, we know that S contains at least one element which is negative. Let's say the smallest element in S is s, and the largest is l, so S = {s, ..., l}.

The range of S is just l - s, and since s is negative, the range of S is larger than l.

The mean of S, on the other hand, cannot be larger than l, the largest element. So we know the mean must be less the range.