Re: jnu 2010

Posted by AJ on May 18, 2012; 7:49am
URL: http://discussion-forum.276.s1.nabble.com/jnu-2010-tp7561215p7565033.html

39.. can be anything...i guess... I mean why we even have this question here...

and 33. ... I am not sure if it's correct but this is what i did...

every X has 5 elements..
and there are 20 such sets...

so,if we just say that these were disjoint.. then there union will have 20*5= 100 elements...

but we are given that every element of union belongs to 10 of the X's .... so that means that union infact has ... 100/10 = 10 elements...

now, coming to Y..

similarly, here union has .. 2*n/4 = n/2 elements...

now both are equal.. that means...  n/2=10 ... so n=20 ..

Does it makes sense..??