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Median

by ranjithreddy.k9 » Sun Jun 12, 2011 7:22 pm
Q. The sum of the integers in list 'A' is the same as the sum of integers in the list 'B' Does List A have less number of integers than List B.

I.The median of set A is more than the median of set B.

II.The average of Integers in set A is more than the average of integers in set B.


a. Statement (I) ALONE is sufficient, but statement (II) alone is not sufficient.
b. Statement (II) ALONE is sufficient, but statement (I) alone is not sufficient.
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
d. Each statement alone is sufficient.
e. Statement (I) and (II) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.



My queation : why is ans not D?
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Source: — Data Sufficiency |
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by cans » Sun Jun 12, 2011 8:17 pm
sum(a)=sum(b)
n(a)<n(b)??
a)med(a)>med(b)
insufficient
a:{0} med=sum=0
b:{-3,-2,-2,7) med=-2 and sum=0
a:{-4,6,3)
b:{5} sum is 5. but n(a)>n(b)

b)avg(a)*n(a) = avg(b)*n(b)
avg(A) > avg(b)
if avg > 0, then n(A)<n(b)
if avg <0, n(a)>n(b)
insufficient

a&b) insufficient
IMO E
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by Spartacus2000 » Mon Jun 13, 2011 10:52 am
Can you provide some examples where the negative avg. affects the answer?

If you think about the average (Avg = Total/no.) as a regular fraction - the value of the fraction increases when the denominator decreases which is what the second statement can be used to imply.

Thus should statement 2 by itself not be enough?
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by ranjithreddy.k9 » Mon Jun 13, 2011 4:25 pm
OA is B
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by phanideepak » Mon Jun 13, 2011 6:46 pm
1: take the sets {2,3,5} median = 3 and {1,1,8} median = 1 n(a) = n(b)

take the sets {2,3,5} median =3 and the set{1,1,1,1,1,1,1,1,1,1} median = 1 n(a) < n(b)

Insuff

2: Sum(a)/n(a) > Sum(b)/n(b) since Sum(a) = Sum(b) We cancel out them and we get n(b) > n(a)

So Sufficient

IMO the answer is B
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