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A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the

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A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 06 Jul 2011, 12:57
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A
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C
D
E

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Question Stats:

61% (01:23) correct 39% (01:11) wrong based on 144 sessions

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A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the reciprocal of the 10th term is x, then x should be in which of the following ranges?

(A) 0.000001 < x < 0.00001
(B) 0.00001 < x < 0.0001
(C) 0.0001 < x < 0.001
(D) 0.001 < x < 0.01
(E) 0.01 < x < 0.1
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 06 Jul 2011, 13:08
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dimri10 wrote:
A sequence of 10 integers is as follows: 2,4,8,16,32.... If the reciprocal of the 10th term is x, then x should be in which of the following ranges?
(A) 0.000001< x <0.00001
(B) 0.00001< x <0.0001
(C) 0.0001< x <0.001
(D) 0.001< x <0.01
(E) 0.01< x <0.1

that's only my OA :lol:


10th term= 2^(10)
Reciprocal = x= 1/2^(10)= 1/1024

x=1/1024 < 1/1000=0.001
x=1/1024 > 1/10000=0.0001 OR 0.0001<x

0.0001<x<0.001

Ans: "C"
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 07 Jul 2011, 16:45
this series in G.P

so 10th term = ar^(n-1)

a=2 r=2 ,n=10

=>10th term = 2^10


reciprocal of 10th term = 1/(2^10) = 1/1024


we know that 1/1000 > 1/1024 > 1/10000

ie 1/1000 > x > 1/10000

0.0001 <x<0.001

Answer is C.
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 07 Jul 2011, 16:57
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\(2^{10} = 1024\)

I think remembering powers of 2 is useful, although it's not hard to quickly calculate this from the pattern.

\(\frac{1}{1024}\), will be a little bit smaller than \(\frac{1}{1000}\) or .001, say .0009

C.
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 08 Jul 2011, 13:01
dimri10 wrote:
A sequence of 10 integers is as follows: 2,4,8,16,32.... If the reciprocal of the 10th term is x, then x should be in which of the following ranges?
(A) 0.000001< x <0.00001
(B) 0.00001< x <0.0001
(C) 0.0001< x <0.001
(D) 0.001< x <0.01
(E) 0.01< x <0.1

that's only my OA :lol:


reciprocal of 10th term is 1/5th term *1/5th term
=1/32*1/32
= .03*.03
=.0009xxx
so itsbigger than .0001 but lesser than .001 :) :)

hope its simple :-D :-D
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 08 Jul 2011, 18:58
10th term is 1024. So the reciprocal is between .001 and .0001. Hence C.
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the  [#permalink]

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New post 01 Mar 2018, 02:56
dimri10 wrote:
A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the reciprocal of the 10th term is x, then x should be in which of the following ranges?

(A) 0.000001 < x < 0.00001
(B) 0.00001 < x < 0.0001
(C) 0.0001 < x < 0.001
(D) 0.001 < x < 0.01
(E) 0.01 < x < 0.1



x = 2^10 = 1024 (Better to remember powers of some numbers)

We know, 1000 < 1024 < 10000

or 1/1000 > 1/1024 > 1/10000

or 0.0001 < x < 0.001

(C)
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Re: A sequence of 10 integers is as follows: 2, 4, 8, 16, 32 ... If the &nbs [#permalink] 01 Mar 2018, 02:56
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