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Mean , Median- of sets -Any Easy method ?

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Mean , Median- of sets -Any Easy method ?

by Chaitanya_1986 » Thu May 05, 2011 12:45 am
Set A, Set B, and Set C each contain only positive integers. If Set A is composed entirely of all the members of Set B plus all the members of Set C, is the median of Set B greater than the median of Set A?
(1) The mean of Set A is greater than the median of Set B.
(2) The median of Set A is greater than the median of Set C.

Any Approach with taking elaborate steps???


OA is E
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Source: — Data Sufficiency |

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by Anurag@Gurome » Thu May 05, 2011 1:56 am
Chaitanya_1986 wrote:Set A, Set B, and Set C each contain only positive integers. If Set A is composed entirely of all the members of Set B plus all the members of Set C, is the median of Set B greater than the median of Set A?
(1) The mean of Set A is greater than the median of Set B.
(2) The median of Set A is greater than the median of Set C.

Any Approach with taking elaborate steps???

OA is E
Question: If {A} = {B} + {C}, then is the Median of {B} > Median of {A}?

(1) Mean of {A} > Median of {B} does not give us any info if Median of {B} > Median of {A} or not, as the mean and median is different for different sets, unless there is any particular condition given (here we only know that Set A, Set B, and Set C each contain only positive integers, which is not a particular condition).
So, (1) is NOT SUFFICIENT.

(2) Median of {A} > Median of {C} again does not give us any info if Median of {B} > Median of {A} or not, as all 3 sets are different from each other.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), we know that Mean of {A} > Median of {B} and Median of {A} > Median of {C}. But again it is NOT SUFFICIENT to find if Median of {B} > Median of {A} or not.

The correct answer is E.
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by ankpta » Tue Nov 08, 2011 12:23 am
IMO :

Let

{A} = (a1,a2,b1,b2,c1,c2}
(B} = (b1,b2)
{C} = (c1,c2)

Now {A} can be arranged as:
1 {a1,a2,b1,b2,c1,c2}
2 {a1,a2,c1,c2,b1,b2}
3 {b1,b2,a1,a2,c1,c2}
4 {b1,b2,c1,c2,a1,a2}
5 {c1,c2,a1,a2,b1,b2}
6 {c1,c2,b1,b2,c1,c2}

1.Condition Median of set A greater than B so b1+b2 > b1+b2 is false condition 1 and 6 cancel out
2.Condition Median of set A greater than C so c1+c2 > c1+c2 is false condition 2 and 4 cancel out

still remain with condition 3 and where position of b1 and b2 not fixed still can't answer the which is greater median .

Pls let me know if my approach is correct

Regards,
Ankit
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by jumsumtak » Mon Dec 05, 2011 10:37 pm
ankpta wrote:IMO :

Let

{A} = (a1,a2,b1,b2,c1,c2}
(B} = (b1,b2)
{C} = (c1,c2)

Now {A} can be arranged as:
1 {a1,a2,b1,b2,c1,c2}
2 {a1,a2,c1,c2,b1,b2}
3 {b1,b2,a1,a2,c1,c2}
4 {b1,b2,c1,c2,a1,a2}
5 {c1,c2,a1,a2,b1,b2}
6 {c1,c2,b1,b2,c1,c2}

1.Condition Median of set A greater than B so b1+b2 > b1+b2 is false condition 1 and 6 cancel out
2.Condition Median of set A greater than C so c1+c2 > c1+c2 is false condition 2 and 4 cancel out

still remain with condition 3 and where position of b1 and b2 not fixed still can't answer the which is greater median .

Pls let me know if my approach is correct

Regards,
Ankit
Set A is composed entirely of all the members of Set B plus all the members of Set C
so no a1, a2..
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