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Depending on whether a, b and c take 0 or 1 , we will have 1,3,10 and 11 as sum in the numerator of the expression .
Answer C
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WE: General Management (Pharmaceuticals and Biotech)
Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1
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06 Nov 2015, 00:43
Nice post, tricky one. IMO C A,B,D,E can be easily eliminated as we have 0 or 1 to choose for values of a,b,c respectively. A ->>> easy to get 1/16 (just setting 0 for a & b) B - >>> setting 0 for a, now we have 2/16 (which is 1/8 reversed to 2/16) we get 3/16 D ->>> Getting (1/2 with 8/16 and b=1, c=0) E - >>> a=b=c=1
Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1
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27 Jan 2018, 09:29
Bunuel wrote:
If \(x = \frac{a}{2} + \frac{b}{2^3} + \frac{c}{2^4}\), 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
Something that may help you in getting to the answer.. All except D are odd/16, so C/16 =1/16....A now work on other two to see if they add up to answer.. a/2+b/8+1/16.... a=b=1... 1/2+1/8+1/16=11/16 ...E a=0 b=1...1/8+1/16 = 3/16...B a=b=0...1/16...A
Ok now 10/16 is nothing but 5/8.. So c is 0... So 1/2+1/8=5/8...D
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73% (01:23) correct 
