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2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?

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2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?  [#permalink]

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New post 25 Oct 2018, 21:23
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A
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D
E

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Question Stats:

73% (01:59) correct 27% (02:44) wrong based on 67 sessions

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Re: 2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?  [#permalink]

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New post 25 Oct 2018, 21:36
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Bunuel wrote:
\(\frac{2^{−6}}{3} + \frac{2^{−7}}{6} + \frac{2^{−8}}{12}\) is how many times bigger than \(2^{−10}\)?


A. 29/21

B. 3

C. 31/6

D. 7

E. 28/3


\(\frac{2^{−6}}{3} + \frac{2^{−7}}{6} + \frac{2^{−8}}{12}\)
= \(\frac{1}{3*2^{6}} + \frac{1}{2*3*2^{7}} + \frac{1}{2^{2}*3*2^{8}}\)
= \(\frac{2^{4}+2^{2}+1}{2^{2}*3*2^{8}}\)
= \(\frac{21}{2^{2}*3*2^{8}}\)
= \(\frac{7}{2^{2}*2^{8}}\)
= \(7*2^{-10}\)

Ans D
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Re: 2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?  [#permalink]

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New post 25 Oct 2018, 22:41
Bunuel wrote:
\(\frac{2^{−6}}{3} + \frac{2^{−7}}{6} + \frac{2^{−8}}{12}\) is how many times bigger than \(2^{−10}\)?


A. 29/21

B. 3

C. 31/6

D. 7

E. 28/3


\(\frac{2^{−6}}{3} + \frac{2^{−7}}{6} + \frac{2^{−8}}{12}\)

\(2^{-8} * (\frac{4}{3} + \frac{2}{6} + \frac{1}{12}\)

\(2^{-8} * \frac{21}{12}\)

We know that \(2^{-10} = \frac{2^{-8}}{4}\)

So our expression is 21/3 (= 7) times more than 2^{-10}

Answer (D)
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Re: 2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?   [#permalink] 25 Oct 2018, 22:41
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