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Posts: 1,015
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
48% (02:04) correct
52%
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wrong
based on 387
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Set X consists of at least 2 members and is a set of consecutive odd integers with an average (arithmetic mean) of 37.
Set Y consists of at least 10 members and is also a set of consecutive odd integers with an average (arithmetic mean) of 37.
Set Z consists of all of the members of both set X and set Y.
Which of the following statements must be true?
I. The standard deviation of set Z is not equal to the standard deviation of set X.
II. The standard deviation of set Z is equal to the standard deviation of set Y.
III. The average (arithmetic mean) of set Z is 37.
A) I only
B) II only
C) III only
D) I and III
E) II and III
Set Y consists of at least 10 members and is also a set of consecutive odd integers with an average (arithmetic mean) of 37.
Set Z consists of all of the members of both set X and set Y.
Which of the following statements must be true?
I. The standard deviation of set Z is not equal to the standard deviation of set X.
II. The standard deviation of set Z is equal to the standard deviation of set Y.
III. The average (arithmetic mean) of set Z is 37.
A) I only
B) II only
C) III only
D) I and III
E) II and III
CAMANISHPARMAR
Set X:-
#elements=2n+1, where \(n\geq{1}\) (since set X contains consecutive odd integers with mean as an odd integer)
Since 37 is the mean, hence set X contains odd numbers of elements with at least 37 as one of the member.
Set Y:-
#elements=2n+1, where \(n\geq{5}\)
Since 37 is the mean, hence set Y contains odd numbers of elements with at least 37 as one of the member.
Set Z:-
Z=X U Y
Now let's evaluate each statement:-
I. SD of set Z will be same as SD of set X when set X and set Y are equal. In all other cases, SD of Z will be different from SD of X. Hence this statement is not true.
II.SD of set Z will be same as SD of set Y when set X and set Y are equal. In all other cases, SD of Z will be different from SD of Y. Hence this statement is not true.
Note:- 2 sets equally spaced , there SD depend on #elements they possess. Greater the no of elements in the set, greater is the SD.
III. Here there are 3 cases, viz,
1. Set Z=Set X=Set Y
2. Set Z= Set X or Set Y
3. Set Z= Set X and Set Y
In the above cases, since both Set X and Y have mean 37. Hence, set Z will always has a mean of 37. hence this statement is correct.
Answer Option. C
PKN
PKN
The Q says that "Set Z consists of all of the members of both set X and set Y". Nothing mentioned here that no repeated members in set Z. So I thought if X =Y, then Z should equals 2X equals 2Y, then statement 2 should be true. Isn't it?
HisHo
Hi HisHo,
First of all, please notice that the question stem is a "MUST BE TRUE". In these type of questions, you need to validate the data at all possible circumstances.
You have found out one possible case where (II) is valid. But there are many cases as highlighted in the explanation where (II) is not valid( When set X and Y are not equal).
Hope it helps.
Posts: 1,348
CAMANISHPARMAR
Not a precise definition of SD, but close enough for the GMAT:
SD = average distance from the mean.
The prompt indicates that X and Y are each composed of consecutive odd integers.
They have the SAME MEAN but could have the SAME NUMBER of terms or a DIFFERENT NUMBER of terms.
Consider an easy case in which X and Y have the same mean and the SAME number of terms:
X = 1, 3, 5
Y = 1, 3, 5
Z = 1, 1, 3, 3, 5, 5
In this case, the average distance from the mean in Z is equal to the average distance from the mean in X, implying that the two sets have the same SD.
The case above illustrates the following:
If X and Y have the same number of terms, then Z and X will have the same SD.
Since Statement I does not have to be true, eliminate A and D.
Consider an easy case in which X and Y have the same mean but a DIFFERENT number of terms:
X = 1, 3, 5, 7, 9
Y = 3, 5, 7
Z = 1, 3, 3, 5, 5, 7, 7, 9
In this case, the values in Z deviate more from the mean than do the values in Y, implying that the two sets do NOT have the same SD.
The case above illustrates the following:
If X and Y have a DIFFERENT number of terms, then Z and Y will NOT have the same SD.
Since Statement II does not have to be true, eliminate B and E.
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We need to recognise few things here before we move on to the Statements.
X >=2. But, for two consecutive odd integers, mean will be even integer. Eg: for 3 and 5 mean is 4, for 39 and 41 mean is 40. This can be extended for even no of consecutive odd integers. So the number of elements in set X will be odd and more than 2, that means, 3,5,7,9,11,13........ with mean of 37.
Same for Set Y but the number of elements ranges from 11,13,15,..... with mean of 37.
Now, Number in set X can be lower of higher than Set Y. because Set Y could be 11 elements and Set X could be 13 elements. Since Standard deviation varies with number of elements in the set. We cannot say which will turn out true between statements I and II so for MBT, both have conditions to turn out not true, so Eliminate I and II.
Now Set Z has all the elements Set X and Y have. But since. Either Set X will be a subset of Y or vice versa or Both could be same sets when both have same number of elements. Then Set Z will also have the same mean. (Statement III proven true). Hence Ans C.
X >=2. But, for two consecutive odd integers, mean will be even integer. Eg: for 3 and 5 mean is 4, for 39 and 41 mean is 40. This can be extended for even no of consecutive odd integers. So the number of elements in set X will be odd and more than 2, that means, 3,5,7,9,11,13........ with mean of 37.
Same for Set Y but the number of elements ranges from 11,13,15,..... with mean of 37.
Now, Number in set X can be lower of higher than Set Y. because Set Y could be 11 elements and Set X could be 13 elements. Since Standard deviation varies with number of elements in the set. We cannot say which will turn out true between statements I and II so for MBT, both have conditions to turn out not true, so Eliminate I and II.
Now Set Z has all the elements Set X and Y have. But since. Either Set X will be a subset of Y or vice versa or Both could be same sets when both have same number of elements. Then Set Z will also have the same mean. (Statement III proven true). Hence Ans C.
CAMANISHPARMAR
for symmetric sets comparing SD is similar to comparing RANGES, so
I. say X has 5 odd numbers [1,3,5,7,9] and Z has 10 odd numbers but the same odd numbers as in set X, so the Range for Z = Range for X thus SD for Z = SD for X but if they have different sets of odd numbers even though they have same number of terms Ranges be different thus SD's can be different ( Not a Must be True Condition )
II. Same logic as above (Not a Must be True Condition)
III. Sum of set X = n * (37)
Sum of set Y = n' * (37)
Sum of set Z = Sum of set X + Sum of set Y = 37 (n + n')
Mean of Z = Sum of set Z / (n + n') = 37 (Must be True)
Please let me know if this is one correct way of solving this
I. say X has 5 odd numbers [1,3,5,7,9] and Z has 10 odd numbers but the same odd numbers as in set X, so the Range for Z = Range for X thus SD for Z = SD for X but if they have different sets of odd numbers even though they have same number of terms Ranges be different thus SD's can be different ( Not a Must be True Condition )
II. Same logic as above (Not a Must be True Condition)
III. Sum of set X = n * (37)
Sum of set Y = n' * (37)
Sum of set Z = Sum of set X + Sum of set Y = 37 (n + n')
Mean of Z = Sum of set Z / (n + n') = 37 (Must be True)
Please let me know if this is one correct way of solving this
CAMANISHPARMAR
ashKing12
ashKing12 Your reasoning about statements I and II is absolutely correct. However, there's a critical flaw in your reasoning for statement III that happens to give you the right answer. Here's why:
The Key Issue with Your Approach to Statement III
You wrote: "Sum of set Z = Sum of set X + Sum of set Y"
This is where the problem lies. This equation only holds if sets X and Y are disjoint (no common elements). However, both X and Y are consecutive odd integers centered at \(37\), which means they must overlap.
Why X and Y Must Overlap
Let me show you why statement III is actually true, but through the correct reasoning:
For consecutive odd integers with mean \(37\):
- Mean \(= a + (n-1)\), where \(a\) is the first term and \(n\) is the number of terms
- For the mean to be \(37\) (odd), \(n\) must be odd (since X needs \(≥ 2\) members, it has \(≥ 3\); Y needs \(≥ 10\) members, it has \(≥ 11\))
- For any fixed odd \(n\): \(a + (n-1) = 37\) gives us \(a = 38 - n\) (unique set!)
Example:
- X with \(3\) terms: \(a = 38 - 3 = 35\) → \({35, 37, 39}\)
- Y with \(11\) terms: \(a = 38 - 11 = 27\) → \({27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47}\)
- Z (union) \(= {27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47} = Y\)
- Mean of Z \(= 37\) ✓
Notice that \(X ⊆ Y\), so \(Z = Y\), and the mean remains \(37\).
The Critical Insight
Since both sets are consecutive odd integers centered at \(37\), set Z (their union) is also a set of consecutive odd integers centered at 37. The union of two symmetric sets centered at the same point produces another symmetric set centered at that point.
Therefore, Mean(Z) = 37 always holds true, but NOT because Sum(Z) = Sum(X) + Sum(Y).
Hope this helps address your doubt!
If you'd like, you can find and practice similar questions here (you'll find a lot of OG questions) - select Statistics under Problem Solving - you can build guided quizzes and customize the difficulty level to solidify your learning for these questions.
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