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1/2^10 + 1/2^11 + 1/2^12 + 1/2^12

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1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 22 May 2020, 07:46
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

77% (01:15) correct 23% (01:25) wrong based on 86 sessions

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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 22 May 2020, 07:56
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Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)


21160


Key property: \(\frac{1}{x^n} = x^{-n}\)
For example, \(\frac{1}{5^8} = 5^{-8}\)


So, we can rewrite our sum as follows: \(2^{-10} + 2^{-11} + 2^{-12} + 2^{-12}\)

Factor out \(2^{-12}\) to get: \(2^{-12}(2^2 + 2^1 + 1 + 1)\)

Evaluate to get: \(2^{-12}(4 + 2 + 1 + 1)\)

Simplify: \(2^{-12}(8)\)

Rewrite as follows: \((2^{-12})(2^3)\)

Apply the product law to get: \(2^{-9}\)

Rewrite as: \(\frac{1}{2^{9}}\)

Answer: C

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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 22 May 2020, 07:51
=> (1/2^10)+(1/2^11)+(1/2^12)+(1/2^12)
Using LCM
=> (1*4/2^10*4)+(1*2/2^11*2)+(1/2^12)+(1/2^12)
=> (4/2^12)+(2/2^12)+(1/2^12)+(1/2^12)
=> (4+2+1+1)/(2^12)
=> 8/2^12
=> 2^3/2^12
=> 1/2^9

Answer is C

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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 22 May 2020, 07:52
Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)


21160


\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} = \frac{1}{2^{10}}*[1+1/2 + 1/4 + 1/8]\)

i.e. \(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} = \frac{1}{2^{10}}*[15/8] ≈ \frac{1}{2^{9}}\)
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 22 May 2020, 08:19
given
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)
can be written as
1/2^10 * ( 1+1/2+1/4+1/4 )
or say
1/2^10 * 8/4 ; 1/2^10 * 2 ;
1/2^9
OPTION C


Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)


21160
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12  [#permalink]

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New post 25 May 2020, 11:05
Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)




First, let’s obtain the common denominator of 2^12. We obtain:

(2^2/2^2)(1/2^10) + (2/2)1/2^11) + 1/2^12 + 1/2^12

2^2/2^12 + 2/2^12 + 1/2^12 + 1/2^12 = 8/2^12 = 2^3/2^12 = 1/2^9

Answer: C
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12   [#permalink] 25 May 2020, 11:05

1/2^10 + 1/2^11 + 1/2^12 + 1/2^12

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