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I'm not sure if I understand the question but I would have to guess E because both answers give no indication of where the mean is, and for all we know there is only one number in set S that is equal to the mean, or there could be 5 or 6.

The range can't be negative, but if all members of a set are negative, the mean is certainly negative. So 1) should be sufficient.

If the median of a set is negative, at least one of the elements of the set is negative. If all of the elements in the set are negative, the range must be larger than the mean, because the range can't be negative. If there is a positive element in the set, then there is a largest positive element- call it x. The mean must be less than x- unless all elements are equal, the mean will certainly be smaller than the largest element. The range must be larger than x, since x is positive and the smallest element is negative. So the range is larger than the mean, and 2) should be sufficient.

Nice question. Answer is D. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Ian, What if the set contains only ONE element>? According to stmt 1 that one no. will be -ive.Then won't the mean and range both be zero? Confused:( _________________

Ian, What if the set contains only ONE element>? According to stmt 1 that one no. will be -ive.Then won't the mean and range both be zero? Confused:(

If the set is something like {-3}, then the range is zero, but the mean is -3. So in that case the range is still bigger than the mean. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

REALLY?Thanks man!Is there a link where such stuff is posted? I always get confused..Lemme try and get it together:

If the set is {-3} Mean: -3 >>I always thought this was Zero.(Mean is the average so -3/1 =-3 ) Median:0 (There is only one element so no middle value) Mode: 0 (Since no value is repeating) Range: 0 (No highest or lowest value) Standard deviation: O (how far are the values dispersed from the mean)

If the set is made up of identical values {1,1,1,1,1} Range= 0,Standard deviation=0

REALLY?Thanks man!Is there a link where such stuff is posted? I always get confused..Lemme try and get it together:

If the set is {-3} Mean: -3 >>I always thought this was Zero.(Mean is the average so -3/1 =-3 ) Median:0 (There is only one element so no middle value) Mode: 0 (Since no value is repeating) Range: 0 (No highest or lowest value) Standard deviation: O (how far are the values dispersed from the mean)

If the set is made up of identical values {1,1,1,1,1} Range= 0,Standard deviation=0

INPUTS?

Median:-3 Mode: 0? 0 cannot be 0 is not in the set. So it could be -3 or none. _________________

REALLY?Thanks man!Is there a link where such stuff is posted? I always get confused..Lemme try and get it together:

If the set is {-3} Mean: -3 >>I always thought this was Zero.(Mean is the average so -3/1 =-3 ) Median:0 (There is only one element so no middle value) Mode: 0 (Since no value is repeating) Range: 0 (No highest or lowest value) Standard deviation: O (how far are the values dispersed from the mean)

If the set is made up of identical values {1,1,1,1,1} Range= 0,Standard deviation=0

INPUTS?

Median:-3 Mode: 0? 0 cannot be 0 is not in the set. So it could be -3 or none.

Could you please explain the median??How is it -3? Mode: hmm..ya i get how that is -3 thanks!! _________________

S1: case-1. S(-1,-1,-1) then Mean=-1 Range=0 so, 0>-1 case-2 S(-1,-2,-3) then mean=-2 range=2 so,2>-2. Sufficient S2: If the median of a set is negative, at least one of the elements of the set is negative or all elements can be nagative,and range cannot be negative so, same cane be explaine as S1. _________________

1. All members of are negative a)if all members are same then range will be 0 which is greater than mean b)if all members are different then range will be > than 0 which is > than mean

either way range is > than mean so choice 1 alone is sufficient

2. The median of set is negative if median is negative then range is definitely positive and greater than mean

Either should be sufficient. (Don't remember which that is.) The mean should be negative for both regardless of the size of the set. So the range would be 0 for only one member (or sets containing the same number) and > 0 for more than one distinct member. So 0 is always > than a negative number.

1) range >=0 and the means are negative - sufficient 2) several cases: a)negative element only -> refer to 1) - sufficient b)more than 1 element - 1) identical elements range=0, median = the element and average = the element => range>mean. 2) non-identical elements range >0, median = negative number, average < the range b.1 and b.2 yields the same result=> 2 is also sufficient

P.S explanation on b.2 as why average is less than range in this case :

If median is negative, that means the left side of the elements ( whether the median is one number or average of two numbers) are less than or equal to the median (or the less of the two numbers) and the right sides are more than or equal to the median( or the bigger of the two numbers). range = Sn-S1 ( S1<0) => range > Sn. Average = Sum (S1, S2, S3....Sn)/ n < Sn. If it is hard to understand, try an example such as { -2, -1, 0, 8} range = 10 >8 and Average = 5/4 < 8

Rephrasing: Is Range [S] > avg[S] S1: Range is always positive as its the difference between extreme points in the set. avg will be -ve as all points are -ve Suff S2: Median being -ve means that mean will be less than range as for eg -32, -31, -30 mean is -31 less then range = 2 -32, -31, 8 mean is -ve which is less then range = 40

Notice that the range of ANY set is more than or equal to zero.

(1) All elements of set S are negative --> the mean of a set with all negative elements is certainly negative so less than its range (which as discussed in always non-negative). Sufficient.

(2) The median of set S is negative --> so there is at least one negative term is the set. Now, consider two cases: A. If all elements in set S are negative then we have the same scenario as above so Range>Mean; B. If not all elements in set S are negative then Range=Largest-Smallest, which will mean that Range>Largest Element (that's because the smallest element in set S is negative. For example consider the following set {-1, -1, 2} --> Range=2-(-1)=3>2). For the same reason the mean will be less than the largest element, so Range>Largest>Mean.

A "range" can never be negative and if all the members of the set are negative, then the mean will definitely be negative. Statement 1 only is sufficient The median only provide no information about the mean nor the range. Statement 2 only is insufficient

A "range" can never be negative and if all the members of the set are negative, then the mean will definitely be negative. Statement 1 only is sufficient The median only provide no information about the mean nor the range. Statement 2 only is insufficient