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(5^55/10^30)(2^30/5^25)=

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Bunuel
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(5^55/10^30)(2^30/5^25)= [#permalink] New post   07 Mar 2016, 03:36
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\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

A. 1
B. 5^30/2^60
C. 5^30/2^30
D. 5^55*2^30/10^30
E. 10^85/50^55

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A
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   07 Mar 2016, 04:39
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Bunuel wrote:
\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

A. 1
B. 5^30/2^60
C. 5^30/2^30
D. 5^55*2^30/10^30
E. 10^85/50^55


Hi
In these Qs we should get the base to prime factors

\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})\)..
\((\frac{5^{55}}{{2^{30}*5^{30}}})(\frac{2^{30}}{5^{25}})=\)..
\((\frac{5^{55-30}}{2^{30}})(\frac{2^{30}}{5^{25}})=\)..
\((5^{25})(\frac{1}{5^{25}})=\).. 1
ans A
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   07 Mar 2016, 04:40
Bunuel wrote:
\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

A. 1
B. 5^30/2^60
C. 5^30/2^30
D. 5^55*2^30/10^30
E. 10^85/50^55


\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})\)
\(=(\frac{5^{55}}{(2*5)^{30}})(\frac{2^{30}}{5^{25}})\)
\(=5^{55}*5^{-30}*5^{-25}*2^{30}*2^{-30}\)
\(=5^{55-30-25}*2^{30-30}\)
\(=5^0*2^0\)
\(= 1\)

The correct answer, therefore, is A
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   10 Jun 2016, 12:34
(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   10 Jun 2016, 14:34
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AbdurRakib wrote:
(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)


Merging topics. Please refer to the discussion above.
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   14 Jun 2016, 00:14
Bunuel wrote:
\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

A. 1
B. 5^30/2^60
C. 5^30/2^30
D. 5^55*2^30/10^30
E. 10^85/50^55


This question uses following formulae of indices:

\(a^m\)*\(a^n\) = \(a^{m+n}\)

\(\frac{1}{{a^m}}\) = \(a^{-m}\)

\(a^m\) / \(a^n\) = \(a^{m-n}\)


\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

\(10^{30}\) can be written as \(5^{30}\) * \(2^{30}\)

Powers of 5 can now be brought together: +55-30-25 = 0

\(5^0\) = 1

Similarly let us arrange power of 2:

-30+30 = 0


\(2^0\) = 1


Answer: 1*1 = 1. A is the answer.
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Re: (5^55/10^30)(2^30/5^25)= [#permalink] New post   14 Jun 2016, 01:46
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AbdurRakib wrote:
(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)


(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) = (5^25/ 2^30) * (2^30/5^25) = 1

Correct Option: A

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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: (5^55/10^30)(2^30/5^25)= [#permalink]
14 Jun 2016, 01:46
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