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Re: Hard statistics [#permalink]
23 Aug 2010, 07:39

Let min be a and max be b. Then range = b-a. 1. R=b-a. If R is added to the set, then the range will not be impacted only if a<=b-a<=b 2a<=b and a>=0. given that numbers are positive, but 2a<=b not given. Hence insuff. 2. New mean < R After adding R new mean becomes (oldM x n + R)/(n+1) < R Hence oldM < R So we know oldM and newM both are less than R. Can't say what the new Range will be from this?

1+2 all positive. And newM, oldM both less than R. Not sure. E

Re: Hard statistics [#permalink]
23 Aug 2010, 07:54

mainhoon wrote:

Let min be a and max be b. Then range = b-a. 1. R=b-a. If R is added to the set, then the range will not be impacted only if a<=b-a<=b 2a<=b and a>=0. given that numbers are positive, but 2a<=b not given. Hence insuff. 2. New mean < R After adding R new mean becomes (oldM x n + R)/(n+1) < R Hence oldM < R So we know oldM and newM both are less than R. Can't say what the new Range will be from this?

1+2 all positive. And newM, oldM both less than R. Not sure. E

Posted from my mobile device

how did you come up with the following? a<=b-a<=b--if you added a on both sides, you would get 2a <=b <=a+b 2a<=b and a>=0.

i pick A. 1) If all the members in set A are positive and R=Max -Min, we get R must be less than the max and greater than the min. therefore the range stays the same for set A.

Re: Hard statistics [#permalink]
23 Aug 2010, 11:16

1

This post received KUDOS

Expert's post

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

Good question. +1.

Let's use the notations proposed by mainhoon: \(a\) - smallest number in the set; \(b\) - largest number in the set; \(r\) - the range, so \(b-a=r\); \(n\) - # of elements in set A; \(m\) - the mean of set A.

The range of new set will NOT increase if \(a\leq{r}\leq{b}\), because new range will still be \(b-a\).

(1) All the numbers in set A are positive --> as all numbers are positive \(r\) can note be more than \(b\), the largest number in the set, so \(r<b\).

But \(r\) can still be less than \(a\) (example \(A=\{5,6\}\), \(r=1\) --> \(A_2=\{1,5,6\}\), \(r_2=5>r=1\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (example \(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(2) The mean of the new set is smaller than R --> \(m_2=\frac{m*n+r}{n+1}<r\) --> \(m<r\) (so R is also more than mean of set A) --> as \(r\) is more than mean of A, then \(r\) can note be less than \(a\), the smallest number in the set, so \(a<r\), (the mean is between the largest and smallest element of the set: \(a\leq{m}\leq{b}\) as \(r>m\), then \(a<r\)).

But \(r\) can still be more than \(b\) (example \(A=\{-5,0\}\), \(r=5\) --> \(A_2=\{-5,0,5\}\), \(r_2=10>r=5\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (\(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(1)+(2) From (1) \(r<b\) and from (2) \(a<r\) --> \(a<r<b\) --> new range will still be \(b-a\), so the answer to the question is NO. Sufficient.

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

While solving this I picked numbers but got stuck in option 2

i tested for the following 2 sets {-1,0,1} and {-1,-2,-3} for both i go range of set A increases .. Hence B, Is there a better and less time consuming way of lookinmg at this ? Thanks a lot.

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

While solving this I picked numbers but got stuck in option 2

i tested for the following 2 sets {-1,0,1} and {-1,-2,-3} for both i go range of set A increases .. Hence B, Is there a better and less time consuming way of lookinmg at this ? Thanks a lot.

Merging similar topics. Please ask if anything remains unclear. _________________

Re: Hard statistics [#permalink]
08 Oct 2012, 21:14

Bunuel wrote:

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

Good question. +1.

Let's use the notations proposed by mainhoon: \(a\) - smallest number in the set; \(b\) - largest number in the set; \(r\) - the range, so \(b-a=r\); \(n\) - # of elements in set A; \(m\) - the mean of set A.

The range of new set will NOT increase if \(a\leq{r}\leq{b}\), because new range will still be \(b-a\).

(1) All the numbers in set A are positive --> as all numbers are positive \(r\) can note be more than \(b\), the largest number in the set, so \(r<b\).

But \(r\) can still be less than \(a\) (example \(A=\{5,6\}\), \(r=1\) --> \(A_2=\{1,5,6\}\), \(r_2=5>r=1\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (example \(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(2) The mean of the new set is smaller than R --> \(m_2=\frac{m*n+r}{n+1}<r\) --> \(m<r\) (so R is also more than mean of set A) --> as \(r\) is more than mean of A, then \(r\) can note be less than \(a\), the smallest number in the set, so \(a<r\), (the mean is between the largest and smallest element of the set: \(a\leq{m}\leq{b}\) as \(r>m\), then \(a<r\)).

But \(r\) can still be more than \(b\) (example \(A=\{-5,0\}\), \(r=5\) --> \(A_2=\{-5,0,5\}\), \(r_2=10>r=5\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (\(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO. Not sufficient.

(1)+(2) From (1) \(r<b\) and from (2) \(a<r\) --> \(a<r<b\) --> new range will still be \(b-a\), so the answer to the question is NO. Sufficient.

Answer: C.

Hi Bunuel - Grt explanation. Many thanks for this.

I have few questions ->

No1 - (2) The mean of the new set is smaller than R.[/quote]

What is is the implication of statement i.e even if the range were smaller than the smallest nos then we can still prove insufficient right, using the below set of values (example \(A=\{5,6\}\), \(r=1\) --> \(A_2=\{1,5,6\}\), \(r_2=5>r=1\)) and in this case answer would be YES but \(a\leq{r}\leq{b}\) is also possible (example \(A=\{1,6\}\), \(r=5\) --> \(A_2=\{1,5,6\}\), \(r_2=5=r=5\)) and in this case answer would be NO.

No2 -> Is is possible to reduce the range by using the same premise "The range of set A is R. A number having a value equal to R is added to set A" From your examples we can see that either range increases or stay the same? Can i decrease?

No3 -> Is there a quick way to come with these nos faster? is it only by sheer practice?

Cheers

gmatclubot

Re: Hard statistics
[#permalink]
08 Oct 2012, 21:14

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