The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
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@ashish100 said: @Stoicalme for ist one:denominator is (a-b)(a+b)now max value of a+b can be till 10 only as 11 won't divide 8!so when a+b=10 then a-b can be 2,4,6,8out of which following sol can be possible : (6,4)(7,3)(8,2)when a+b=9 then a-b can be 1,3,5,7here 0 possible solutions as max power of 3 can be two onlywhen a+b=8 a-b can be 2,4,6so (5,3)(6,2)possible solutionswhen a+b=7 and a-b can be 1,3,5possible sol are :(4,3)(5,2)when a+b=6 and a-b can be 2,4so sol can be :(4,2)when a+b =5 and a-b can be 1,3so sol can be 3,2 when a+b=4 then a-b can be 2so sol can be nothingIn all getting 9 solutions....kahan miss kar raha hoon bhai log