Thanks Mahesh, I corrected it.

26 Sep 2015 - 7:13pm

as they r in proportion i.e     a/b=c/d so ad=bc---------(i)

now solving     (a-b)(a-c)/a = a^2-ac-ab+bc/a from eqn (i)

(a-b)(a-c) = a^2-ac-ab+ad

​taking a common i.e a(a-c-b+d)/a

cancelling a with 1/a gives a+d-b-c i.e OPTION B

20 Sep 2014 - 11:27pm

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