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

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Intern
Joined: 14 Oct 2010
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A is a set containing 7 different numbers. B is a set contai  [#permalink]

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14 Oct 2010, 08:40
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60% (01:25) correct 40% (01:34) wrong based on 370 sessions

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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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Joined: 02 Sep 2009
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14 Oct 2010, 09:08
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2
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;

Hope it's clear.
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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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12 Jul 2013, 15:05
1
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
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Re: A is a set containing 7 different numbers. B is a set contai  [#permalink]

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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.

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

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25 Sep 2018, 23:16
Distilling the problem down to set notation, you can see that Set B = {n1,n2,n3,n4,n5,n6}

Set A contains everything in Set B, but with an additional number. In essence, Set A = {n1,n2,n3,n4,n5,n6,N}

The additional number, N, is the only thing that can be manipulated in this problem, so that is what you should naturally focus on. The problem then asks you to find out which of the statements given CANNOT be true. Turning this around from the viewpoint of the Testmaker, your job is then to try to discover a condition where each statement IS true.

Quote:
Answer choice “A” tells you that the range of set A is equal to the range of set B. Naturally, this could work if N were somewhere in the middle of the set, meaning that it wouldn’t affect the range. Answer choice “A” COULD BE true, and therefore isn’t the right answer.

Quote:
Answer choice “B” tells you that the mean of set A is greater than the mean of set B. Naturally, an unusually large value for N would increase the average of set A. Answer choice “B” COULD BE true, and therefore isn’t the right answer.

Quote:
Answer choice “C” tells you that the range of set A is less than the range of set B. If the additional number, N, were somewhere in the middle of the set, then it wouldn’t affect the range and set A and set B would have equal ranges. If the number, N, were bigger or smaller than the other numbers, it would increase the range for set A. It could never make the range smaller. Answer choice “C” can NEVER be true, and therefore must be the right answer.

Quote:
Answer choice “D” tells you that the mean of set A is equal to the mean of set B. This is a bit trickier to find an example for, but there is one case when this happens. Remember: adding the average of a set to a set doesn’t change the average. If the additional number, N, were equal to the average of set B, then both sets would still have the same average. Answer choice “D” COULD BE true, and therefore isn’t the right answer.

Quote:
Answer choice “E” tells you that the median of set A is equal to the median of set B. This, too, is a bit tricky, until you remember how to calculate median. Remember that the median of a set with an even number of elements (like set B) is equal to the average of the two middle numbers; in this case, the median is not contained in the set. If the additional number, N, were equal to the median of set B, it would also fall as the middle number of set A, and the medians would be the same. Answer choice “E” COULD BE true, and therefore isn’t the right answer.

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

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26 Sep 2018, 01:32
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

OA: C

Let Set A: { -1,0,1,2,3,4,5} $$\quad Range_{A} = 5-(-1) = 5+1=6$$
Let Set B1: { -1,0,1,2,3,4} $$\quad Range_{B1} = 4-(-1) = 4+1 = 5$$
Let Set B2: { -1,0,1,2,3,5} $$\quad Range_{B2} = 5-(-1) = 5+1 = 6$$
Let Set B3: { -1,0,1,2,4,5} $$\quad Range_{B3} = 5-(-1) = 5+1 = 6$$
Let Set B4: { -1,0,1,3,4,5} $$\quad Range_{B4} = 5-(-1) = 5+1 = 6$$
Let Set B5: { -1,0,2,3,4,5} $$\quad Range_{B5} = 5-(-1) = 5+1 = 6$$
Let Set B6: { -1,1,2,3,4,5} $$\quad Range_{B6} = 5-(-1) = 5+1 = 6$$
Let Set B7: { 0,1,2,3,4,5} $$\quad Range_{B7} = 5-(0) = 5-0 = 5$$

The range of A is never less than the range of B
Re: A is a set containing 7 different numbers. B is a set contai   [#permalink] 26 Sep 2018, 01:32
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