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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, 20:23
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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, 20:36
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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, 21:41
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]

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New post 21 May 2020, 15:01
VeritasKarishma

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

Kindly, explain how you get 21/3?
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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 24 May 2020, 15:14
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


Simplifying, we have:

2^(-6) / 3 + (2^1)(2^(-8)) / 6 + (2^2)(2^(-10)) / 12

2^(-6) / 3 + 2^(-8) / 3 + 2^(-10) / 3

2^(-10) * (2^4 + 2^2 + 1) / 3

2^(-10) * 21 / 3

2^(-10) * 7

We see that the given expression is 7 times bigger 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]

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New post 04 Jun 2020, 04:22
VeritasKarishma wrote:
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)


Hi, I have a doubt, possibly a very stupid one,
If x=7y
isn't x 6 time more than y
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Re: 2−6/3 + 2−7/6 + 2−8/12 is how many times bigger than 2^−10?   [#permalink] 04 Jun 2020, 04:22

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

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