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# If the variables, X, Y, and Z take on only the values 10, 20, 30, 40

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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only
(B) Y only
(C) Z only
(D) X and Y
(E) X and Z

Sol: Ans is E

Consider data for Variable X

We see that it is symmetric about 40 and hence the distribution is equally spaced and therefore Mean =Median
We can also spread the distribution as
10,20,20,30,30,30,40,40,40,40,50,50,50,60,60,70.. Strike out from extremes and we get Median= 40
And the Mean will be (40*4+10*1+20*2+30*3+50*3+60*2+70*1)/16 or 640/16= 40

Similar symmetry can be observed for Z distribution as well.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Isnt it too dificult to culculate !

I was going for x only.. bt looking at the z ..it seems it will have same median and mean..what is easiest way to solve
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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When the distribution is symmetric, mean equals median. Thus, X and Z must be in the answer. There is no choice with X, Y, and Z, so there is no need to check Y. The answer is E.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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For X: The distribution is symmetric and more populated at mean. So for this mean = median
For Y: The distribution is neither symmetric nor populated at mean. So for this mean is not equal to median
For Z: The distribution is symmetric but more populated at ends. So mean is shifted towards right side. Thus mean is not equal to median.

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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Answer = (E) X and Z

Looking at the charts, see that X & Z are symmetric.

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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
As per the 3 figures given in the question , it can be clearly seen that for graphs X and Z mean is equal to median
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
try to estimate, this is what worked for me:
both X and Z looks similar and approx. gets us to the same mean and median
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Bunuel, VeritasPrepKarishma, EgmatQuantExpert please share the theory of these types of problem.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Bunuel, VeritasPrepKarishma, EgmatQuantExpert please share the theory of these types of problem.

The theory of this question is the same as the theory of mean and median. It uses no new concepts. In both X and Z, the numbers are symmetrical about the centre. So the sum of the deviation of numbers on the left of the centre is same as the sum of the deviation of the numbers on the right.
This post explains finding mean using deviations: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/

If the set of numbers is 10, 15, 15, 20, 25, 25, 30, the mean will simply be 20. Notice the symmetry around the centre, 20. The median will obviously be 20 too.

The case with X and Z is exactly the same.

Y has a random sequence. There is no symmetry about the centre and hence the mean will not be the same as the median.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Bunuel wrote:

If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only
(B) Y only
(C) Z only
(D) X and Y
(E) X and Z

Responding to a pm:
Go back to the basics. What is mean? It's the average, the single value that can represent all the values. How do you find it? You multiply each value by its frequency, add them all up and divide by the sum of the frequencies.
What is the mean of: 10, 20, 30 - we know it is 20
What is the mean of: 10, 10, 20, 30, 30 - again 20. Why? Because if there is one value '10' which is 10 less than 20, then there is also a value '30' which is 10 more than 20. So effectively, both 10 and 30 give us two 20s.
Similarly, here X is: 10, 20, 20, 30, 30, 30, 40, 40, 40, 40, 50, 50, 50, 60, 60, 70
40 is the mean because we have three 30s and three 50s to balance out. We have two 20s and two 60s to balance out and we have a 10 and a 70 to balance out again.
What is median? Median is the middle value when you arrange the numbers in increasing/decreasing order. We can see that 40 will be the middle value too since there are equal number of total elements on both sides of 40. We have 6 elements smaller than 40 and 6 elements greater than 40. Hence median = 40.

Hence, mean = median for X.

We can reason out the same thing for Z too in exactly the same way.

The only possible answer is (E). I don't have to worry about Y.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
In addition to the above logic mentioned for symmetry around the center that conveys for equal mean and median, one can also quickly analyze to find the mean by using the benchmark value of 40 in X and Z. 40 being a benchmark value in X can easily be seen to get a mean of 40 by calculating the positive and negative deviations o either side. Same logic can be extended to see that the benchmark value of 40 in Z also gives equal positive and negative deviations on either side.
The answer to this question is figure X and Z.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
Bunuel wrote:

If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only
(B) Y only
(C) Z only
(D) X and Y
(E) X and Z

Responding to a pm:
Go back to the basics. What is mean? It's the average, the single value that can represent all the values. How do you find it? You multiply each value by its frequency, add them all up and divide by the sum of the frequencies.
What is the mean of: 10, 20, 30 - we know it is 20
What is the mean of: 10, 10, 20, 30, 30 - again 20. Why? Because if there is one value '10' which is 10 less than 20, then there is also a value '30' which is 10 more than 20. So effectively, both 10 and 30 give us two 20s.
Similarly, here X is: 10, 20, 20, 30, 30, 30, 40, 40, 40, 40, 50, 50, 50, 60, 60, 70
40 is the mean because we have three 30s and three 50s to balance out. We have two 20s and two 60s to balance out and we have a 10 and a 70 to balance out again.
What is median? Median is the middle value when you arrange the numbers in increasing/decreasing order. We can see that 40 will be the middle value too since there are equal number of total elements on both sides of 40. We have 6 elements smaller than 40 and 6 elements greater than 40. Hence median = 40.

Hence, mean = median for X.

We can reason out the same thing for Z too in exactly the same way.

The only possible answer is (E). I don't have to worry about Y.

What an awsm explanation!

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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
Top Contributor
Bunuel wrote:

If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only
(B) Y only
(C) Z only
(D) X and Y
(E) X and Z

The data distributions for X and Z are symmetric about 4. So the mean and median of the $$X$$ and $$Z$$ are equal $$=40$$

The answer is $$E$$
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
KarishmaB wrote:
Bunuel, VeritasPrepKarishma, EgmatQuantExpert please share the theory of these types of problem.

The theory of this question is the same as the theory of mean and median. It uses no new concepts. In both X and Z, the numbers are symmetrical about the centre. So the sum of the deviation of numbers on the left of the centre is same as the sum of the deviation of the numbers on the right.
This post explains finding mean using deviations: https://www.gmatclub.com/forum/veritas- ... eviations/

If the set of numbers is 10, 15, 15, 20, 25, 25, 30, the mean will simply be 20. Notice the symmetry around the centre, 20. The median will obviously be 20 too.

The case with X and Z is exactly the same.

Y has a random sequence. There is no symmetry about the centre and hence the mean will not be the same as the median.

hi KarishmaB, Thanks for the great replies across GMAT club. They have been very helpful.

My question: Is it guaranteed that if there isn't a symmetry in the histogram, then mean ≠ median ?
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
sriharsha4444 wrote:
KarishmaB wrote:
Bunuel, VeritasPrepKarishma, EgmatQuantExpert please share the theory of these types of problem.

The theory of this question is the same as the theory of mean and median. It uses no new concepts. In both X and Z, the numbers are symmetrical about the centre. So the sum of the deviation of numbers on the left of the centre is same as the sum of the deviation of the numbers on the right.
This post explains finding mean using deviations: https://www.gmatclub.com/forum/veritas- ... eviations/

If the set of numbers is 10, 15, 15, 20, 25, 25, 30, the mean will simply be 20. Notice the symmetry around the centre, 20. The median will obviously be 20 too.

The case with X and Z is exactly the same.

Y has a random sequence. There is no symmetry about the centre and hence the mean will not be the same as the median.

hi KarishmaB, Thanks for the great replies across GMAT club. They have been very helpful.

My question: Is it guaranteed that if there isn't a symmetry in the histogram, then mean ≠ median ?

No, it is not essential that mean will not be equal to median if there is no symmetry.

Consider: 2, 3, 6, 7, 12

Mean = median = 6

But here the options are such that we can ignore Y.
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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Re: If the variables, X, Y, and Z take on only the values 10, 20, 30, 40 [#permalink]
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