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D01-06

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D01-06  [#permalink]

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New post 16 Sep 2014, 00:11
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

85% (01:04) correct 15% (01:15) wrong based on 172 sessions

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Math Expert
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Re D01-06  [#permalink]

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New post 16 Sep 2014, 00:11
Official Solution:

What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}}\)?

A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)


We just have to simplify the expression:
\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}} = \frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4}} = \frac{\frac{1}{2^3}}{\frac{1}{2^3} * \frac{1}{2^4}} = 2^4 = 16\)


Answer: A
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Re: D01-06  [#permalink]

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New post 08 Jun 2018, 03:56
Quote:
Official Solution:

What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}}\)?

A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)


We just have to simplify the expression:
\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}} = \frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4}} = \frac{\frac{1}{2^3}}{\frac{1}{2^3} * \frac{1}{2^4}} = 2^4 = 16\)


Answer: A


Thank you for elaborating.

I somehow have issues with such problems on a frequent basis.

Does breaking the respective fraction down into the respective prime and exponent represent the best go to solution?
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Re: D01-06  [#permalink]

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New post 08 Jun 2018, 06:38
Arro44 wrote:
Quote:
Official Solution:

What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}}\)?

A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)


We just have to simplify the expression:
\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2}} * 2^{-4}} = \frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4}} = \frac{\frac{1}{2^3}}{\frac{1}{2^3} * \frac{1}{2^4}} = 2^4 = 16\)


Answer: A


Thank you for elaborating.

I somehow have issues with such problems on a frequent basis.

Does breaking the respective fraction down into the respective prime and exponent represent the best go to solution?


In many cases yes, because in this case we break a number into its building blocks which makes manipulating a lot easier.
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Re: D01-06  [#permalink]

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New post 21 Aug 2018, 10:29
everything just cancel out by itself and we are left with 1 upon 1by 16 which after reversal becomes 16.
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Re: D01-06   [#permalink] 21 Aug 2018, 10:29
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