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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

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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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New post 05 Nov 2015, 12:24
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A
B
C
D
E

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  25% (medium)

Question Stats:

78% (01:55) correct 22% (02:06) wrong based on 136 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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New post 05 Nov 2015, 12:59
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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New post 05 Nov 2015, 18:57
1
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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New post 05 Nov 2015, 22:54
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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New post 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
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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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New post 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

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] 27 Jan 2018, 09:29
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