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If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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05 Nov 2015, 11:24
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73% (01:24) correct 27% (01:09) wrong based on 79 sessions
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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05 Nov 2015, 11:59
Bunuel wrote: If x = a/2 + b/2^3 + 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 x = a/2 + b/2^3 + c/2^4 = 8a/16 + 2b/16 + c/16 (A) 1/16 when a=b=0, c=1 (B) 3/16 when a = 0, b=c=1 (D) 10/16 when a=b=1, c=0 (E) 11/16 when a=b=c=1 (C) 5/16 there is no value of a, b, c (0 or 1) Ans: C



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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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05 Nov 2015, 17:57
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x= a/2 + b/2^3 + c/2^4 = a/2 + b/8 + c/16 = (8a + 2b + c)/16 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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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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05 Nov 2015, 21:54
x cannot be 5/16
Ans:c



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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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05 Nov 2015, 23:43
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



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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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27 Jan 2018, 08:29
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 Ans C
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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, 08:29






