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If the variables, X, Y, and Z take on only the values [#permalink]

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06 Nov 2011, 14:33

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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

I was wondering do we need to order the values to find mean and median in X,Y and Z?

Z = 40,30,20,10,20,30,40

So do we order from lowest value to highest?

If so we get 10,20,20,30,30,40,40

Thus mean = 190/7 = 27.14

and median = 30

and so mean is not equal to median ???

Please explain when do we order the values and when do we not order the values ?

Also, OG explanation doesnt take frequencies into consideration. It just looks at the symmetry and says mean= median for X and Z whcih doesnt make any sense

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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10 Nov 2011, 22:24

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siddhans wrote:

Attachment:

Mean_Median.png

I was wondering do we need to order the values to find mean and median in X,Y and Z?

Z = 40,30,20,10,20,30,40

So do we order from lowest value to highest?

If so we get 10,20,20,30,30,40,40

Thus mean = 190/7 = 27.14

and median = 30

and so mean is not equal to median ???

Please explain when do we order the values and when do we not order the values ?

Also, OG explanation doesnt take frequencies into consideration. It just looks at the symmetry and says mean= median for X and Z whcih doesnt make any sense

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. _________________

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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23 Jul 2014, 00:26

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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

The frequency distributions for both X and Z are symmetric about 40, which means that both X and Z have mean = median = 40.

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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08 Nov 2011, 03:53

loser wrote:

To solve this question you need to check for symmetry, which can be visualized in x and z.

why dont we consider the frequencies given in this problem too ??? whats the use then of the frequencies given?how do we know we shouldnt consider the frequencies and re-arrange them ?

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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09 Nov 2011, 01:46

siddhans wrote:

loser wrote:

To solve this question you need to check for symmetry, which can be visualized in x and z.

why dont we consider the frequencies given in this problem too ??? whats the use then of the frequencies given?how do we know we shouldnt consider the frequencies and re-arrange them ?

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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11 Nov 2011, 02:42

VeritasPrepKarishma wrote:

siddhans wrote:

Attachment:

Mean_Median.png

I was wondering do we need to order the values to find mean and median in X,Y and Z?

Z = 40,30,20,10,20,30,40

So do we order from lowest value to highest?

If so we get 10,20,20,30,30,40,40

Thus mean = 190/7 = 27.14

and median = 30

and so mean is not equal to median ???

Please explain when do we order the values and when do we not order the values ?

Also, OG explanation doesnt take frequencies into consideration. It just looks at the symmetry and says mean= median for X and Z whcih doesnt make any sense

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.

Wow, i had no idea that if X has 4 shaded squares it should be taken 4 times. To me it looked like that X = 40 (if its shaded at 40) and not 4 times 40 .

How do we know if we need to take 4 times 40 and not just 40 ....? Since at 10 X is shaded once and so according to that logic at 40 it should be 10+10+10+10 and not 40+40+40+40?

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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14 Nov 2011, 10:19

Expert's post

siddhans wrote:

Wow, i had no idea that if X has 4 shaded squares it should be taken 4 times. To me it looked like that X = 40 (if its shaded at 40) and not 4 times 40 .

How do we know if we need to take 4 times 40 and not just 40 ....? Since at 10 X is shaded once and so according to that logic at 40 it should be 10+10+10+10 and not 40+40+40+40?

They mention in the question that it is a frequency distribution so anyway there is no doubt about what the shaded squares mean. The diagram tells you how frequently the value appears in the data e.g. in X, 2 squares are shaded above 60 which means that 60 appears twice. _________________

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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09 Jan 2012, 20:37

VeritasPrepKarishma wrote:

siddhans wrote:

Wow, i had no idea that if X has 4 shaded squares it should be taken 4 times. To me it looked like that X = 40 (if its shaded at 40) and not 4 times 40 .

How do we know if we need to take 4 times 40 and not just 40 ....? Since at 10 X is shaded once and so according to that logic at 40 it should be 10+10+10+10 and not 40+40+40+40?

They mention in the question that it is a frequency distribution so anyway there is no doubt about what the shaded squares mean. The diagram tells you how frequently the value appears in the data e.g. in X, 2 squares are shaded above 60 which means that 60 appears twice.

Sorry to bring back an old thread.

Can someone explain this more thoroughly?? I understand the MEDIAN part, it's the middle number. But the MEAN part (the average) makes no sense.

AVERAGE = SUM OF NUMBERS/frequency

Average of 1,3,5,7,9,11 = (36)/6 = 4 ;right...

SO FOR EXAMPLE of X, why is not [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16 16 is the frequency no?? If we divide all these numbers by 16, we get some funky decimal number.

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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09 Jan 2012, 21:01

Expert's post

vq35boi wrote:

Sorry to bring back an old thread.

Can someone explain this more thoroughly?? I understand the MEDIAN part, it's the middle number. But the MEAN part (the average) makes no sense.

AVERAGE = SUM OF NUMBERS/frequency

Average of 1,3,5,7,9,11 = (36)/6 = 4 ;right...

SO FOR EXAMPLE of X, why is not [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16 16 is the frequency no?? If we divide all these numbers by 16, we get some funky decimal number.

Incorrect calculation. When you do [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16, you get 40 as expected. You can figure it out without doing this calculation. Go back to the explanation I provided above of what average means. It helps you deduce that average in this case has to be 40 without calculating. (or check out a standard Arithmetic book to understand this.) The formula you have given just helps you calculate average if you need to calculate it. _________________

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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10 Jan 2012, 05:41

@VeritasPrepKarishma, dont u think that we can see the right answer even visually. I mean a close look to the pics can reveal, that the white regions are evenly spaced from the median 40. or my approach is more intuitive and I have found the right answer just by luck? _________________

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Re: If the variables, X, Y, and Z take on only the values [#permalink]

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11 Jan 2012, 00:33

Expert's post

LalaB wrote:

@VeritasPrepKarishma, dont u think that we can see the right answer even visually. I mean a close look to the pics can reveal, that the white regions are evenly spaced from the median 40. or my approach is more intuitive and I have found the right answer just by luck?

You are required to see it visually and arrive at the answer. The symmetry shows that the mean and the median will both lie in the middle i.e. at 40. The calculations show the reason the mean will lie in the center in case of symmetrical data. Ideally, you should not do these calculations during the exam. _________________

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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18 Mar 2012, 07:15

VeritasPrepKarishma wrote:

LalaB wrote:

@VeritasPrepKarishma, dont u think that we can see the right answer even visually. I mean a close look to the pics can reveal, that the white regions are evenly spaced from the median 40. or my approach is more intuitive and I have found the right answer just by luck?

You are required to see it visually and arrive at the answer. The symmetry shows that the mean and the median will both lie in the middle i.e. at 40. The calculations show the reason the mean will lie in the center in case of symmetrical data. Ideally, you should not do these calculations during the exam.

Great Explanation ! however there is one minor subset that we are all missing here. Symmetry results in "mean equals median" however symmetry ( the kinda visual symmetry that we are referring to here) is not a prerequisite.

Consider the following dist that is a slight modification of the Y graph....

so while X and Z can be easily chosen. A simple residual analysis(sum of variation of value over/below mean) of Y wouldnt hurt.( in this case it is not req coz there is only one option with x and z).

Am i thinking on the right lines or have i totally lost it (john nash style) since i have been doin quants for 8 hrs straight....

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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19 Mar 2012, 10:22

Expert's post

emailsatz wrote:

VeritasPrepKarishma wrote:

LalaB wrote:

@VeritasPrepKarishma, dont u think that we can see the right answer even visually. I mean a close look to the pics can reveal, that the white regions are evenly spaced from the median 40. or my approach is more intuitive and I have found the right answer just by luck?

You are required to see it visually and arrive at the answer. The symmetry shows that the mean and the median will both lie in the middle i.e. at 40. The calculations show the reason the mean will lie in the center in case of symmetrical data. Ideally, you should not do these calculations during the exam.

Great Explanation ! however there is one minor subset that we are all missing here. Symmetry results in "mean equals median" however symmetry ( the kinda visual symmetry that we are referring to here) is not a prerequisite.

Consider the following dist that is a slight modification of the Y graph....

so while X and Z can be easily chosen. A simple residual analysis(sum of variation of value over/below mean) of Y wouldnt hurt.( in this case it is not req coz there is only one option with x and z).

Am i thinking on the right lines or have i totally lost it (john nash style) since i have been doin quants for 8 hrs straight....

You are right that symmetrical data implies mean = median but it is not a prerequisite. But in this question, we see that for X and Z, "mean = median" so it is not possible that for Y "mean = median" because there is no option giving all three (X, Y and Z). So answer has to be (D) and that is why we can ignore Y. Had there been an option which included Y too, we would have to find out whether this relation holds for Y. _________________

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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21 Mar 2012, 08:09

True! This problem doesn't warrant an analysis of Y .But on the 700 plus range it is safer to do a quick residual analysis on similar problems that include the asymmetrical option too.Thanks a ton for your response !

Quote:

You are right that symmetrical data implies mean = median but it is not a prerequisite. But in this question, we see that for X and Z, "mean = median" so it is not possible that for Y "mean = median" because there is no option giving all three (X, Y and Z). So answer has to be (D) and that is why we can ignore Y. Had there been an option which included Y too, we would have to find out whether this relation holds for Y.[/quote]

Re: If the variables, X, Y, and Z take on only the values [#permalink]

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