Re: DSE 2009 and 2011 contradiction

Posted by rahul on Jun 19, 2012; 7:28am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2009-and-2011-contradiction-tp7578178p7578185.html

y1+ y2=6
(a, 5)+ min(a,2)=6
so possible value of "a " will be in in between 2 & 5
so use a=3
then 3+2<6
now a=4,
4+2=6 satisfy the condition
hence (4,2)