Is the standard deviation of the numbers X, Y and Z equal to the standard deviation of 10, 15 and 20?

(1) Z - X = 10. No info about y. Not sufficient.
(2) Z - Y = 5. . No info about x. Not sufficient.

(1)+(2) From above x = z - 10 and y = z - 5, so the set in ascending order is {z-10, z-5, z}. Now, if we add or subtract a constant to each term in a set the standard deviation will not change. Adding 20-z to each term in the set we get {10, 15, 20}. So, the standard deviation of {z-10, z-5, z} is equal to that of {10, 15, 20}. Sufficient.

Answer: C.

Hope it's clear.
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Sweet Trick to solve the question, very helpful!

Another way to look at SD is to think in terms of a number line. SD calculates the dispersion of numbers from the mean. The SD of two sets will be the same if the relative placement of numbers from the respective means is the same.

This is what 10, 15 and 20 will look like on a number line
10 .... 15 .... 20
(15 is the mean and 10 and 20 are 5 steps away from the mean. Each dot is a number between 10 and 15 and between 15 and 20)

(1) Z - X = 10
This is what Z and X will look like on the number line
X ......... Z

(2) Z - Y = 5
This is what Z and Y will look like on the number line
Y .... Z

Together, their relative placement on the number line looks like this:
X .... Y .... Z

This matches the placement of 10, 15 and 20 and hence the SD will be the same in the two cases.

For more on number line concepts of SD, check: http://www.veritasprep.com/blog/2012/06 ... deviation/
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1.
From the information we know that the gap between Z and X is 10 so we can think of any number with that gap...
{5,Y,15} or {10,Y,20} or {50,Y,60}, etc. These sets are similar to the given {10,15,20} in such away that the first and last term are of a distance of 10.

Notice that the middle number of {10,15,20} is 15 which is equal to the average = 20+10+15/3 = 15. Now, we do not know the middle number or Y or {X,Y,Z}. If Y is equal to the average then it will have an SD equal to the SD of {10,15,20}. If Y is not equal to the average, then our SD will be greater.

INSUFFICIENT!


2. From the information we know that Y and Z are of 5 away from each other {X,15,20} or {X,16,21}, etc. These sets are similar to {10,15,20} in terms of the distance of 2nd to the last term. But, we need to know X to know how spread out are the numbers. If X -Y is 5 then the SD will be the same. If not then the SD will not be the same.

INSUFFICIENT!

Together:
{X,Y,Z} = {i, i+5, i+10} SD is the same with {10,15,20} where i=10: {10, 10+5, 10+10}

Answer: C

well I thought the same way and was going to mark C
But I stopped thinking of another case

Y....Z.........X
I.e this case satisfies the 2 conditions difference between Y and Z is 5 and difference between Z and X is 10, will SD be same in this case too?
Sounds a bit stupid, but need to know why this approach is incorrect?

Z - X = 10 implies that Z is greater than X by 10 which means Z MUST be to the right of X on the number line. It doesn't matter whether Z and X are both positive, both negative or one positive one negative. You cannot put Z to the left of X on the number line and still have Z - X = 10. This is the reason using number line is a good idea because it gives you a lot of clarity.
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Also, SD of 10, 15, 20 will not be the same as SD of Y....Z.........X (e.g. 5, 10, 20). The distance of the numbers from the mean is not the same in the two cases.

SD of 10, 15, 20 will be the same as SD of 20, 25, 30 or of 41, 46, 51 or of -16, -11, -6 etc.
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I feel like this problem was easier for me to solve when having this in mind "Is x, y, and Z -- 10, 15, and 20".

And then just on C putting z = 20 and calculate. However I realized that it is unnecessary since you can tell that C is sufficient.

Is the 700-level rating accurate?

I would say 650 - 700. Note that there are certain complications:

1. The concept of SD is not very intuitive to many people which makes this question hard. Once you understand it, you feel its simple.
2. X, Y and Z are not given to be positive so subtraction puts people off sometimes since they feel they have to account for positive as well as negative numbers. Its all in the perception.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the standard deviation of the numbers X, Y and Z equal to the standard deviation of 10,15 and 20?

(1) Z - X = 10
(2) Z - Y = 5

Looking at the original condition, there are 3 variables (x,y,z) and in order to match the number of variables and equations, we need 3 equations. However, only 2 equations are given from the 2 conditions, so there is high chance (E) will be our answer
Looking at the conditions together,
In 10,15,20, there are differences of 5 between 10&15, 15&20, but a difference of 10 between 10 and 20.
According to x,y,z, there is a difference of 5 between x and y, and also 5 between y and z, and 10 between z and x. The standard deviation becomes identical, and therefore the conditions are sufficient, making the answer (C)

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Shouldn't we consider negative numbers for x, y & z unless specified that they are only positive? If you consider -ve numbers, x = -20, y = -5, z = -10 satisfies both the conditions. The distance between the numbers is not the same as it would have been with positive numbers. Shouldn't the answer be 'both statements not sufficient'?

If y = -5 and z = -10 then Z - Y = -5, not 5.

Also, most of the solutions above deal with the problem conceptually, which includes all cases.
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Target question: Is the standard deviation (SD) of the numbers X, Y and Z equal to the standard deviation of 10,15 and 20?

IMPORTANT CONCEPT: In order for the SD of X, Y and Z to be equal to the SD of {10, 15, 20}, X, Y and Z must have the same dispersion as {10, 15, 20}
That is, if X, Y and Z are arranged in ascending order, the 2nd value must be 5 greater than the 1st value, and the 3rd value must be 5 greater than the 2nd value.
So, for example, the following sets will have the same standard deviation as {10, 15, 20}:
{1, 6, 11}
{8.3, 13.3, 18.3}
{-8, -3, 2}
Etc

Statement 1: Z - X = 10
No information about Y
So, statement 1 is NOT SUFFICIENT

Statement 2: Z - Y = 5
No information about X
So, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that Z - X = 10, which means Z is 10 greater than X
Statement 2 tells us that Z - Y = 5, which means Z is 5 greater than Y
So, we can conclude that Z is the biggest number.

If we take the equation Z - X = 10 and subtract the equation Z - Y = 5, we get -X + Y = 5, which is the same as Y - X = 5
This tells us that Y is 5 greater than X

So, we now know that Z is 5 greater than Y and Y is 5 greater than X
So, {X, Y, Z} has the exact same dispersion as {5, 10, 20}, which means {X, Y, Z} has the same standard deviation as {5, 10, 20}
Since we can answer the [color=blue]target question with certainty, the combined statements are SUFFICIENT

Answer:

RELATED VIDEO

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Hi All,

While this question requires some specific knowledge about Standard Deviation, you don't actually have to do much math to solve it. To start, it's worth noting that the GMAT will NEVER ask you to calculate the Standard Deviation of a group using the S.D. formula, so that is NOT what this question is actually about.

We're asked if the S.D. of three numbers (X, Y and Z) is the SAME S.D. as the one for the numbers 10, 15 and 20. This is a YES/NO question.

The numbers 10, 15 and 20 are 'evenly spaced' numbers that differ by 5. To have the same S.D. as this group, another group must ALSO have evenly spaced numbers that differ by 5. For example, (0, 5, 10) and (1, 6, 11) would have the same S.D. as (10 15. 20).

1) Z - X = 10

This Fact fits part of the pattern that we're looking for, but we don't know the relative value of Y.
Fact 1 is INSUFFICIENT

2) Z - Y = 5

This Fact also fits part of the pattern that we're looking for, but we don't know the relative value of X.
Fact 2 is INSUFFICIENT

Combined, we know....
Z - X = 10
Z - Y = 5

This means that Z is the largest value, that Z is 5 greater than Y, and that Z is 10 greater than X. By extension, Y would then be 5 greater than X. This is an exact match for the 'spread' created by (10, 15, 20), so the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Bunuel Do you mean sd of 40 , 45 , 50 will be same as 10,15,20?
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Yes, if we add or subtract a constant to each term in a set the standard deviation will not change. If we subtract 30 from each term in {40 , 45 , 50} we get {10, 15, 20}, so the SD of these two sets are the same.
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Check these posts for basics of standard deviation:
https://www.veritasprep.com/blog/2012/0 ... deviation/
https://www.veritasprep.com/blog/2012/0 ... n-part-ii/
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.


Since we have 3 variables and 0 equations, E is most likely to be the answer and so we should consider 1) & 2) first.

The distance 10 and 20 is 10 and the distance 15 and 20 is 5.
The distance X and Z is 10 and the distance Y and Z is 5.
Since the distributions of two data sets are same, their standard deviations are same.
Both conditions 1) & 2) are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1)
Since we don't know anything about Y, this is not sufficient.

Condition 2)
Since we don't know anything about X, this is not sufficient.

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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