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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
GMATPrepNow wrote:
Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x}

If x < 0, and the standard deviation of set A = the standard deviation of set B, what is the value of x?

A) -2
B) -3
C) -4
D) -5
E) -6


mean of set A ; A = {22, 27, 11, 12, 2} ; 74/5; 14.8
and SD along mean ; 12.8,2.8,3.8,0,7.2,12.2
similarly at x=-6 we get the mean of set B ; 40-6/5 ; 34/5 ; 6.8
SD along mean is 12.8,2.8,3.8,0,7.2,12.2
IMO E
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
Total range lower than mean and bigger than mean set A=12+13=25, set B= 9+10+x=25 additional 6 needed, hence -6
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
The easiest way to solve this question is to line up the numbers of the sets in increasing order. That is start set A from 2 and start set B from x. You will notice that when you line them up the difference between each number of the respective set lines up. 11 is one way from 12, same as 3 is one away from 4. 12 is 10 away from 22, same as 4 is 10 away from 14 and so on.

Now to make sure that the standard deviation is the same, all we need to do is ensure that the difference between x and 3 is the same the difference between 2 and 11. That is x is 9 less than 3. Every other position is already set so the only movement to standard deviation can come from the value of x. To standardise that we insert -6 therefore smoothing out the distribution to ensure the same standard deviation. It more logic than solving complex formulae.
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
Expert Reply
Mean gets affected if the data in the set are increased or decreased by adding or subtracting the same number respectively. But S.D. remains the same.

Set A= {22, 27, 11, 12, 2}
Set B= {14, 19, 3, 4, x}

There is a difference of '8' between each term of Set A and Set B. Therefore, 2- (x) = 8

=> x = -6

Answer E
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Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
I solved it this way. Let me know if this is the right approach to solve this. Bunuel

Since, SD <= Range/2

Range of Set A = 27-2 = 25
so, SD of set A <= 12.5
Given, SD of set A = SD of set B = 12.5

Range of Set B = 2*SD = 19-(-x) <= 25 => x <= 25-19 <= 6

since, x is negative, therefore, x <= -6 (E)
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
BrentGMATPrepNow I arranged the set in ascending order and instead of considering the mean I thought about the median. Could you tell me how does this thinking work out despite not considering the mean??

Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x}

A = (2,11,12,22,27
B = (x,3,4,14,19)

Now we know that SD is the distance from the mean. BUT I yet considered the median to get an idea.

Here I noticed that A has a net displacement of 36 (from 12) and that B has a net displacement of 26 (from4. So X needs to compensate for that extra 10 (since the SD of the two sets are the same)

Hence x = -6
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
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Hoozan wrote:
BrentGMATPrepNow I arranged the set in ascending order and instead of considering the mean I thought about the median. Could you tell me how does this thinking work out despite not considering the mean??

Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x}

A = (2,11,12,22,27
B = (x,3,4,14,19)

Now we know that SD is the distance from the mean. BUT I yet considered the median to get an idea.

Here I noticed that A has a net displacement of 36 (from 12) and that B has a net displacement of 26 (from4. So X needs to compensate for that extra 10 (since the SD of the two sets are the same)

Hence x = -6


For this particular question, that approach works perfectly.
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
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Re: Set A = {22, 27, 11, 12, 2}, and set B: {14, 19, 3, 4, x} [#permalink]
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