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Section - 32 Qs - 9

Expert replies
by camitava » Sat Oct 13, 2007 5:18 am
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.
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Correct me If I am wrong


Regards,

Amitava
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Source: — Problem Solving |

by Suyog » Sat Oct 13, 2007 6:53 am
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.
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by camitava » Sat Oct 13, 2007 7:01 am
Thanks Suyog! Got it... Thanks a lot once again...
Correct me If I am wrong


Regards,

Amitava
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