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15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=

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15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post Updated on: 31 May 2017, 20:51
1
5
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (01:52) correct 29% (02:27) wrong based on 140 sessions

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\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\)

A. 2^7

B. 2^8

C. 2^9

D. 15(2^7)

E. 15 (2^8)

Originally posted by pmklings on 18 Sep 2015, 19:33.
Last edited by Bunuel on 31 May 2017, 20:51, edited 1 time in total.
Renamed the topic and edited the question.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 18 Sep 2015, 20:38
I think you are asking the value of

\(15/(2^{-5}+2^{-6}+2^{-7}+4^{-4})\)

We can write it as

\(15 / (2^{-5}+2^{-6}+2^{-7}+2^{-8})\) = \(15/ 2^-{5}*(1 + 2^{-1} + 2^{-2} + 2^{-3})\) = \(15*2^5 /(15/8)\)
On simplification, we get \(2^5*2^3\)

Hence the answer\(2^8\)

Please make sure that you type the question properly.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 18 Sep 2015, 20:57
thank you for the quick response. I am unsure how i wrote the equation in incorrectly. Could you please inform me of my error?
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 18 Sep 2015, 21:05
pmklings wrote:
thank you for the quick response. I am unsure how i wrote the equation in incorrectly. Could you please inform me of my error?

I am not sure as to what you are referring to. If you can write down what you have solved, I might be able to help you in a better way.

For a solution to the problem, you can refer to the above post.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 18 Sep 2015, 21:09
you said there was a problem with the way i wrote the question. i am trying to see how it is wrong.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 31 May 2017, 20:46
How did you get (15/8) in the last step? Until that step everything else looks straighforward, I am sure I am missing some silly small step.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 31 May 2017, 20:54
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 31 May 2017, 21:06
2
Kp19 wrote:
How did you get (15/8) in the last step? Until that step everything else looks straighforward, I am sure I am missing some silly small step.


Formatted and added few more steps.

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\)

A. 2^7

B. 2^8

C. 2^9

D. 15(2^7)

E. 15 (2^8)

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\frac{15}{2^{-5}+2^{-6}+2^{-7}+2^{-8}}=\)

Factor out 2^(-5) from the denominator:

\(=\frac{15}{2^{-5}(1+2^{-1}+2^{-2}+4^{-3})}=\frac{15*2^5}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}=\frac{15*2^5}{\frac{8}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}=\frac{15*2^5}{\frac{15}{8}}=15*2^5*\frac{8}{15}=2^8\).

Answer: B.

Hope it's clear.
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 01 Jun 2017, 07:17
Bunuel wrote:
Kp19 wrote:
How did you get (15/8) in the last step? Until that step everything else looks straighforward, I am sure I am missing some silly small step.


Formatted and added few more steps.

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\)

A. 2^7

B. 2^8

C. 2^9

D. 15(2^7)

E. 15 (2^8)

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\frac{15}{2^{-5}+2^{-6}+2^{-7}+2^{-8}}=\)

Factor out 2^(-5) from the denominator:

\(=\frac{15}{2^{-5}(1+2^{-1}+2^{-2}+4^{-3})}=\frac{15*2^5}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}=\frac{15*2^5}{\frac{8}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}=\frac{15*2^5}{\frac{15}{8}}=15*2^5*\frac{8}{15}=2^8\).

Answer: B.

Hope it's clear.


Hi Bunuel,

I have a question. I flipped the original equation and made it (2^5+2^6+2^7+2^8)/15 and got 2^5 as my final answer. Could you please help me what exactly I did wrong? Are we not allowed to take reciprocals here at the beginning?
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 01 Jun 2017, 07:25
1
csaluja wrote:
Bunuel wrote:
Kp19 wrote:
How did you get (15/8) in the last step? Until that step everything else looks straighforward, I am sure I am missing some silly small step.


Formatted and added few more steps.

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\)

A. 2^7

B. 2^8

C. 2^9

D. 15(2^7)

E. 15 (2^8)

\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\frac{15}{2^{-5}+2^{-6}+2^{-7}+2^{-8}}=\)

Factor out 2^(-5) from the denominator:

\(=\frac{15}{2^{-5}(1+2^{-1}+2^{-2}+4^{-3})}=\frac{15*2^5}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}=\frac{15*2^5}{\frac{8}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}=\frac{15*2^5}{\frac{15}{8}}=15*2^5*\frac{8}{15}=2^8\).

Answer: B.

Hope it's clear.


Hi Bunuel,

I have a question. I flipped the original equation and made it (2^5+2^6+2^7+2^8)/15 and got 2^5 as my final answer. Could you please help me what exactly I did wrong? Are we not allowed to take reciprocals here at the beginning?


No, this is wrong. \(\frac{1}{2^{-1}+2^{-2}}=\frac{4}{3}\) but \(2^{1}+2^{2}=6\).
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 05 Jun 2017, 03:19
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Easiest way as per me, change 2^-5 to 1/2^5 which is 1/32, do the same to other terms as we so we'll get,

15/ (1/32+1/64+1/128+1/256) = 15/ [(8+4+2+1)/256] = 15/15/256 = 256 = 2^8.

Took only 48 sec to get the answer by this method. Option B is correct.

Kudos if you like my method. Thanks :)
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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New post 06 Jun 2017, 16:41
pmklings wrote:
\(\frac{15}{2^{-5}+2^{-6}+2^{-7}+4^{-4}}=\)

A. 2^7

B. 2^8

C. 2^9

D. 15(2^7)

E. 15 (2^8)


Let’s first simplify the denominator of the fraction:

2^-5 + 2^-6 + 2^-7 + 4^-4

1/2^5 + 1/2^6 + 1/2^7 + 1/4^4

1/32 + 1/64 + 1/128 + 1/256

8/256 + 4/256 + 2/256 + 1/256 = 15/256

So, we have:

15/(15/256) = 256 = 2^8.

Alternative Solution:

Let’s multiply the numerator and the denominator of the fraction by 2^8:

(2^8 * 15)/(2^8 * (2^-5 + 2^-6 + 2^-7 + 2^-8)

(2^8 * 15)/(2^3 + 2^2 + 2^1 + 2^0)

(2^8 * 15)/(8 + 4 + 2 + 1)

(2^8 * 15)/15 = 2^8

Answer: B
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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=  [#permalink]

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Re: 15/(2^(-5)+2^(-6)+2^(-7)+4^(-4))=   [#permalink] 25 Mar 2019, 09:07
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