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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 02 Sep 2015, 22:28
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

82% (01:39) correct 18% (02:39) wrong based on 77 sessions

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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 02 Sep 2015, 22:58
1
Bunuel 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) \(15 (2^{8})\)


Kudos for a correct solution.


Denominator=1/2^5+2^6+2^7+2^8=1/2^3+2^2+2+1/2^8
Fraction becomes=15*2^8/15=2^8
Answer B
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15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 03 Sep 2015, 00:02
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Simplifying the denominator of this fraction in necessary to calculate,as below

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

=\(\frac{15}{2^-5+2^-6+2^-7+2^-8}\)
=\(\frac{15}{2^-5(1+2^-1+2^-2+2^-3)}\)
=\(\frac{15}{2^-5(1+1/2+1/4+1/8)}\)
=\(\frac{15}{2^-5(15/8)}\)
=\(\frac{15}{2^-5(15/2^3)}\)
=\(\frac{15}{2^-8(15)}\)
=\(\frac{1}{2^-8}\)
=2^8

So,the Correct answer is B
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 03 Sep 2015, 08:27
1
Bunuel 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) \(15 (2^{8})\)


Kudos for a correct solution.


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

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

= \(\frac{15}{{2^{-5}({{1+2^{-1}+2^{-2}+2^{-3}})}\)

= \(\frac{(2^5)(15)}{(1 + 1/2 + 1/4 + 1/8)}\)

=\(\frac{(2^5)(15)}{(15/8)}\)

= \(8(2^5)\)

= \((2^3)(2^5)\)

= \(2^8\)

Answer:- B
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 04 Sep 2015, 14:30
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\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\) =?

The denominator becomes 2^-5(15/8)

15*(8/15)*2^-5
=2^8

Answer:
(B) \(2^{8}\)
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 06 Sep 2015, 07:58
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Solution:

Denominator = \(2^{-5} + 2^{-6} + 2^{-7} + 2^{-8}\) = \(2^{-8}[2^{3} + 2^{2} + 2 + 1]\) = \(2^{-8}(8+4+2+1)\) = \(2^{-8}(15)\)

Therefore, ans = \(\frac{(15)}{(2^{-8}(15))}\) = \(2^{8}\)

Option B
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 07 Sep 2015, 03:45
Bunuel 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) \(15 (2^{8})\)


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

When you face exponents and addition, you’ll almost always need to factor out common exponential terms so that you can multiply, which is the operation that lends itself to more flexibility with exponent rules. Here, you can first take \(4^{-4}\) and express it as \((2^{2})^{-4}\), which equals \(2^{-8}\), so that you have all common bases in the denominator.

Then, in order to multiply, factor out a \(2^{-5}\) from each term to get: \(2^{-5} (1 + 2^{-1} + 2^{-2} + 2^{-3})\). The \(2^{-5}\) is now a multiplicative term, which you can express also as \(\frac{1}{(2^{5})}\), allowing you to flip it to the numerator, which becomes \(2^{5}* 15\).

Back to the denominator, you can express \((1 + 2^{-1} + 2^{-2} + 2^{-3})\) as \(1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}\), which equals \(1 + \frac{7}{8} = \frac{15}{8}\).

Putting the entire fraction together, the fraction is \(\frac{15(2^{5})}{(\frac{15}{8})}\). The 8 can be expressed as \(2^{3}\), again providing a common exponential base, making the fraction \(\frac{15(2^{5})}{\frac{15}{2^{3}}}\). Dividing by a fraction is the same as multiplying by its reciprocal, leaving \(\frac{15(2^{5}) * 2^{3}}{15}\). The 15s cancel, leaving \(2^{5} * 2^{3}\) which equals \(2^{8}\), answer choice B.
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}  [#permalink]

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New post 16 Mar 2019, 19:33
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}   [#permalink] 16 Mar 2019, 19:33
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