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R is a set containing 8 different numbers. S is a set containing 7 di
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12 Aug 2021, 07:42
12 Aug 2021, 07:42
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R is a set containing 8 different numbers. S is a set containing 7 different numbers, all of which are members of R. Which of the following statements CANNOT be true?
(A) The range of R is less than the range of S.
(B) The mean of R is greater than the mean of S.
(C) The range of R is equal to the range of S.
(D) The mean of R is less than the mean of S.
(E) The mean of R is equal to the mean of S.
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) The range of R is less than the range of S.
(B) The mean of R is greater than the mean of S.
(C) The range of R is equal to the range of S.
(D) The mean of R is less than the mean of S.
(E) The mean of R is equal to the mean of S.
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
Re: R is a set containing 8 different numbers. S is a set containing 7 di
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12 Aug 2021, 17:04
12 Aug 2021, 17:04
Solution:
Lets say the value of the set are as below:
R={0, 1, 2, 3, 4, 5, 6, 7}
S= {1, 2, 3, 4, 5, 6, 7} or {0, 2, 3, 4, 5, 6, 7} or {0, 1, 2, 4, 5, 6, 7} or......
You can add or remove any number in set S
(A) The range of R is less than the range of S.---> This is not possible as either the two sets can have equal range or the range of set R can be greater than the range of set S. Consider removing 7 the range decreases. As we cannot add another number which is greater than the number in set R this option cannot be true.- Correct
(B) The mean of R is greater than the mean of S.--->Possible- If I remove 7 from set S and find the mean I get it as 3 which is less than the mean of set R which is 3.5---OUT
(C) The range of R is equal to the range of S.---> Possible- Set R & S starting from 0 and ending with 7-OUT
(D) The mean of R is less than the mean of S.---> This is surely possible, consider the above set of S from 1-7 which has a mean of 4 and set R has a mean of 3.5- OUT
(E) The mean of R is equal to the mean of S.-->Although this is not possible in the above set we can try another set
R= {1, 2, 3, 4, 6, 8, 10, 14}
S= {1, 2 , 3, 4, 8, 10, 14}
Both of the above set have a mean of 6- OUT
IMO- A
Hope this helps!
Lets say the value of the set are as below:
R={0, 1, 2, 3, 4, 5, 6, 7}
S= {1, 2, 3, 4, 5, 6, 7} or {0, 2, 3, 4, 5, 6, 7} or {0, 1, 2, 4, 5, 6, 7} or......
You can add or remove any number in set S
(A) The range of R is less than the range of S.---> This is not possible as either the two sets can have equal range or the range of set R can be greater than the range of set S. Consider removing 7 the range decreases. As we cannot add another number which is greater than the number in set R this option cannot be true.- Correct
(B) The mean of R is greater than the mean of S.--->Possible- If I remove 7 from set S and find the mean I get it as 3 which is less than the mean of set R which is 3.5---OUT
(C) The range of R is equal to the range of S.---> Possible- Set R & S starting from 0 and ending with 7-OUT
(D) The mean of R is less than the mean of S.---> This is surely possible, consider the above set of S from 1-7 which has a mean of 4 and set R has a mean of 3.5- OUT
(E) The mean of R is equal to the mean of S.-->Although this is not possible in the above set we can try another set
R= {1, 2, 3, 4, 6, 8, 10, 14}
S= {1, 2 , 3, 4, 8, 10, 14}
Both of the above set have a mean of 6- OUT
IMO- A
Hope this helps!
gmatclubot
Re: R is a set containing 8 different numbers. S is a set containing 7 di [#permalink]
12 Aug 2021, 17:04
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