Hi,

Let us use statement 2 right away. (Statement 1 is not sufficient)

Let us look at this step by step.

Consider just the top (1st) scorer of Dr. Adam's class. He has to be paired with one guy from Dr. B's class whose marks are more than his. Mark this guy from Dr. B's class. He can no longer be utilized in another pair.

Consider the 2nd scorer of Dr. A's class. He has to be paired with one of higher scorers in Dr. B's class and who is still 'unmarked'. Once he has been selected, mark him also. He can no longer be utilized in another pair.

Consider the 3rd scorer of Dr. A's class. He is the median guy in Dr. A's class. He has to be paired with one of the higher scorers in Dr. B's class and who is still 'unmarked'. One would be mistaken to assume that this new guy is the median of Dr. B's class. He need not be the median. (A case in point - If everybody in Dr. B's class gets more than Dr. A's class members).

The contention is that if there is some third person in Dr. B's class who has scored more than the median score of Dr. A's class; then the median of Dr. B's class has to be greater than the median Dr. A's class. Hence, 2 is sufficient.

Hope this helps. Thanks.

1) statment 1, it has only given the range of the both groups.Brown's max and min ranges of the scroes, which is 50 and 90 ,similarly for Adam's class which are 40 to 90.
But there can be any values between the lower and higher ranges..

for Brown's class we may assume {50,55,55,70, 90}
for Adam's class we may assume {40,50,60,70,80}
here the median of the Brown's ( 3rd term in acending or decending order) is less than that of Adam's...

another example,
we may take another example of Brown's and Adam's classes

for Brown's , it could be {50,60,65,72,90}
Adam's class , it may be {40,45,48,70,80}

Here, the median of Brown's class is more than that Of Adam's.

Hence, the STATMENT 1 is not sufficient.

2) statment 2, we take this qualitatively..
as it says in evey pair of two people , higher scorer of each pair is of brown's student.
make a chart

(A1,B1)=(40,50)
(A2,B2)=(41,58)
(A3,B3)=(79,80)
(A4,B4)=(65,66)
(A5,B5)=(80,90)

hence, Set of Brown's ={50,58,66,80,90} and set of adam's={40,41,65,78,80}
so brown's median is greater than that of Adam's

or It could be

(A1,B1)=(78,90)
(A2,B2)=(40,50)
(A3,B3)=(62,70)
(A4,B4)=(80,81)
(A5,B5)=(55,56)

putting into ascending or decedning order of both the sets , the brown's median is always greater than Adam's for any such arrnagments , so we can say the answer as "NO".
hence statment 2 is sufficient .

Answer :B
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