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D01-06

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Bunuel
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D01-06 [#permalink] New post   16 Sep 2014, 00:11
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Question Stats:

87% (01:20) correct 13% (01:57) wrong based on 177 sessions
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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. \(\frac{1}{16}\)
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A
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Bunuel
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Re D01-06 [#permalink] New post   16 Sep 2014, 00:11
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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. \(\frac{1}{16}\)


Simplify the given expression by expressing all values as powers of 2:

\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} } = \)

\(=\frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4} } = \)

\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)

\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)

\(= \frac{1}{\frac{1}{2^4} } = \)

\(= 2^4=\)

\(= 16\)


Answer: A
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Re: D01-06 [#permalink] 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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Bunuel
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Re: D01-06 [#permalink] New post   08 Jun 2018, 06:38
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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] New post   24 Oct 2019, 17:03
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I think this is a high-quality question and I agree with explanation. Reclassify this to Exponents, not Arithmetic please
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Re: D01-06 [#permalink] New post   11 Jul 2023, 12:34
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: D01-06 [#permalink]
11 Jul 2023, 12:34
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