jnu

[k]

2011-q63  
2011-q62

[Sinistral]

if I got the right questions, then soln. is as under:
62)
look at the first option:
x< xy/2 < y
dividing by x throughout (wont change inequality since x>0)
1<y/2<y/x
2<y<(2/x)y    ---1

now look at the given inequality:
0<x<2<y       --- 2
note that 2<y is being satisfied in 1
see inequality (2): x<2 ==> 2/x  > 1
hence see the right inequality of (1):
y<      (2/x)   *y
y< (a no. > 1)*y
so true
hence option a
check to see that all other options are false.

63)
SUM OF DIGITS =13
i.e. 2+1+3+x+y=13  ==> x+y=7
now remainder is less than 10 when divided by 100 ==> xy is of the form 01 or 02 or 03 ...or 09 ===> x=0 ==> y=7 (since x+y=7)
hence option (b)