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A is a set containing 7 different numbers. B is a set contai

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A is a set containing 7 different numbers. B is a set contai [#permalink]

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New post 14 Oct 2010, 08:40
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A is a set containing 7 different numbers. B is a set containing 6 different numbers, all of which are members of A. Which of the following statements CANNOT be true?

A) The range of A is equal to the range of B
B) The mean of A is greater than the mean of B
C) The range of A is less than the range of B
D) The mean of A is equal to the mean of B
E) The median of A is equal to the median of B
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Re: Help with range [#permalink]

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New post 14 Oct 2010, 09:08
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The range of a set is the difference between the largest and smallest elements of a set.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

The arithmetic mean (average) of a set equals the sum of all of the elements in a set divided by the number of elements in that set.

Consider the set A to be {-3, -2, -1, 0, 1, 2, 3} --> \(mean=median=0\) and \(range=3-(-3)=6\).

A. The range of A = range of B --> remove 0 from set A, then the range of B still would be 6 --> could be true;

B. The mean of A > the mean of B --> remove 3, then the mean of B would be negative -0.5 so less than 0 --> could be true;

C. The range of A < the range of B --> the range of a subset cannot be more than the range of a whole set: how can the difference between the largest and smallest elements of a subset be more than the difference between the largest and smallest elements of a whole set --> NEVER TRUE.

D. The mean of A = the mean of B --> remove 0 from set A, then the mean of B still would be 0 --> could be true;

E. The median of A = the median of B --> again remove 0 from set A, then the median of B still would be 0 --> could be true;

Answer: C.

Hope it's clear.
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Re: Help with range [#permalink]

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New post 19 May 2011, 01:40
Good concept.

C it as range can be at max = range of B but cant be less than B.
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Re: A is a set containing 7 different numbers. B is a set contai [#permalink]

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New post 12 Jul 2013, 15:05
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Jumphi97 wrote:
A is a set containing 7 different numbers. B is a set containing 6 different numbers, all of which are members of A. Which of the following statements CANNOT be true?

A) The range of A is equal to the range of B
B) The mean of A is greater than the mean of B
C) The range of A is less than the range of B
D) The mean of A is equal to the mean of B
E) The median of A is equal to the median of B


hi,

only C is not possible:

let say set A has : a== b== c== d== e== f== g==>these are 7 elements in ascending order(just think they are on the number line starting from left)
range =g-a(distance from a to g)
now for B we have select 6 elements from A...and have to leave 1 element.
Now for making B YOU HAVE TO remove any one....no wyou can clearly see if you remove any of the one b/c/d/e/f===>range is g-a==>same as set A's range.
if you remove g==>range is f-a==>clearly distance is less than g-a
if you remove a ==>range is g-b==>clearly distance is less than g-a

hence you can see range of set B is always less than or equal to range of set A

hope i made my point clear :-D
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Re: A is a set containing 7 different numbers. B is a set contai [#permalink]

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New post 21 Jan 2018, 20:04
Jumphi97 wrote:
A is a set containing 7 different numbers. B is a set containing 6 different numbers, all of which are members of A. Which of the following statements CANNOT be true?

A) The range of A is equal to the range of B
B) The mean of A is greater than the mean of B
C) The range of A is less than the range of B
D) The mean of A is equal to the mean of B
E) The median of A is equal to the median of B


Since all of the number in set B are in set A, and A has 1 more number than set B, it would be impossible for the range of set A to be any less than set B.

Answer: C
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Re: A is a set containing 7 different numbers. B is a set contai   [#permalink] 21 Jan 2018, 20:04
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