Please refer the qs below -
9. If the above fig is true, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT
(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16
And OA is C
But I solved the qs in following way -
if a,b and c = 1, x = 11/16 now the numerator will always be odd as a, b and c can be either 0 or 1. Getting me what I want to explain? Pls guys help me to point out my fault here.
Section - 32 Qs - 9
This topic has expert replies
I agree that numerator will always be odd.
But just consider a case where c = 0 and a, b = 1
the eqn will be
a/2 + b/2^3 + c/2^4
= 1/2 + 1/8 + 0/16
= 4/8 + 1/8
= 5/8
Here the numerator is still odd but the denominator is not 16 and will not be 16 always.
so this ans 5/8 becomes 10/16 (Choice 4)
and hence x could not be 5/16.
Ans C.
But just consider a case where c = 0 and a, b = 1
the eqn will be
a/2 + b/2^3 + c/2^4
= 1/2 + 1/8 + 0/16
= 4/8 + 1/8
= 5/8
Here the numerator is still odd but the denominator is not 16 and will not be 16 always.
so this ans 5/8 becomes 10/16 (Choice 4)
and hence x could not be 5/16.
Ans C.