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Re: 0.0001^(1/2) [#permalink]
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\(\sqrt{0.00001}\)

=\(\sqrt{10^{-5}*1}\)

=\(\sqrt{10^{-6}*10}\) --> Easier to get \(10^{-6}\) out of the square root

=\(10^{-3}\)*\(\sqrt{10}\)

=\(\frac{1}{10^3}\)*\(\sqrt{10}\)

=\(\frac{1}{1000}\)*\(\sqrt{10}\)

=\(\sqrt{10}\)/1000
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Re: 0.0001^(1/2) [#permalink]
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the answer is attached. this question was tricky because the exponents and roots component. thanks for your responses, you helped me to better understand this problem.
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0.0001^(1/2) [#permalink]
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lalania1 wrote:
\(\sqrt{0.00001}\) ?

A) \(\sqrt{10}\)/100

B) 1/100

C) \(\sqrt{10}\)/1000

D) 1/1000

E) \(\sqrt{10}\)/10000


Rule #1: √(a/b) = (√a)/(√b)
Rule #2: √(ab) = (√a)(√b)


So, √0.00001 = √(1/100,000)
= (√1)/(√100,000) [applied rule #1]
= 1/(√100,000)

Let's take a closer look at √100,000
√100,000 = √(10,000 x 10)
= (√10,000)(√10) [applied rule #2]
= 100(√10)

So, 1/(√100,000) = 1/100√10
Check the answer choices...not there.
Looks like we need to "fix" (rationalize) the denominator (for more on this, check the video below)
Multiply top and bottom by √10 to get: (√10)(1)/(√10)(100√10)
Simplify: (√10)/1000

Answer: C

Originally posted by BrentGMATPrepNow on 24 Nov 2016, 11:41.
Last edited by BrentGMATPrepNow on 28 Oct 2022, 06:50, edited 1 time in total.
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Re: 0.00001^(1/2) = [#permalink]
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Bunuel wrote:
\(\sqrt{0.00001}\) =


A) \(\frac{\sqrt{10}}{100}\)

B) \(\frac{1}{100}\)

C) \(\frac{\sqrt{10}}{1000}\)

D) \(\frac{1}{1000}\)

E) \(\frac{\sqrt{10}}{10000}\)


OA: C
\(\sqrt{0.00001}\) = \(\sqrt{10^{-5}}=10^{-2}*10^{-\frac{{1}}{{2}}}= \frac{1}{{100\sqrt{10}}}\)
Multiplying numerator and denominator by \(\sqrt{10}\)
\(\frac{{1}}{{100\sqrt{10}}}\)\(*\)\(\frac{{\sqrt[]{10}}}{{\sqrt[]{10}}}\) \(=\frac{{\sqrt[]{10}}}{1000}\)
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0.00001^(1/2) = [#permalink]
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Approach and Working:

\(\sqrt{(0.00001)}= \sqrt{(1/10^5)}= \sqrt{(10/10^6)}= \frac{\sqrt{10}}{ 10^3}= \frac{\sqrt{10}}{ 1000}\)

Hence, the correct answer is option C.

Answer: C
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0.0001^(1/2) [#permalink]
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Expert Reply
lalania1 wrote:
\(\sqrt{0.00001}\) ?


A) \(\frac{\sqrt{10}}{100}\)

B) \(\frac{1}{100}\)

C) \(\frac{\sqrt{10}}{1000}\)

D) \(\frac{1}{1000}\)

E) \(\frac{\sqrt{10}}{10000}\)


Here is how one can approach it: We see that the options do not have any decimals. So let's get rid of the decimal.

\(\sqrt{0.00001} = \sqrt{\frac{1}{100000}} = \frac{1}{100\sqrt(10)} \)

Now note that it doesn't match any option but some options have root 10 in the numerator. So multiply and divide by sqrt(10) to get

\(\frac{1*\sqrt(10)}{100\sqrt(10) * \sqrt(10)} = \frac{\sqrt(10)}{1000}\)

Answer (C)

Check out this video on exponents:

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Re: 0.00001^(1/2) = [#permalink]
Expert Reply
Bunuel wrote:
\(\sqrt{0.00001}\) =


A) \(\frac{\sqrt{10}}{100}\)

B) \(\frac{1}{100}\)

C) \(\frac{\sqrt{10}}{1000}\)

D) \(\frac{1}{1000}\)

E) \(\frac{\sqrt{10}}{10000}\)


\(\sqrt{0.00001}........................\sqrt{\frac{1}{100000}}=\frac{1}{100}*\sqrt{\frac{1}{10}}.........................\frac{\sqrt{10}}{1000}\)

C
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Re: 0.0001^(1/2) [#permalink]
BrentGMATPrepNow wrote:
lalania1 wrote:
\(\sqrt{0.00001}\) ?

A) \(\sqrt{10}\)/100

B) 1/100

C) \(\sqrt{10}\)/1000

D) 1/1000

E) \(\sqrt{10}\)/10000


Rule #1: √(a/b) = (√a)/(√b)
Rule #2: √(ab) = (√a)(√b)


So, √0.00001 = √(1/100,000)
= (√1)/(√100,000) [applied rule #1]
= 1/(√100,000)

Let's take a closer look at √100,000
√100,000 = √(10,000 x 10)
= (√10,000)(√10) [applied rule #2]
= 100(√10)

So, 1/(√100,000) = 1/100√10
Check the answer choices...not there.
Looks like we need to "fix" (rationalize) the denominator (for more on this, check the video below)
Multiply top and bottom by √10 to get: (√10)(1)/(√10)(100√10)
Simplify: (√10)/1000
Answer:

RELATED VIDEOS





Hi BrentGMATPrepNow, √ 1 * 10^-4 = 10^-2 = 1/100. What have I missed here? Thanks Brent.
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Re: 0.0001^(1/2) [#permalink]
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Kimberly77 wrote:
Hi BrentGMATPrepNow, √ 1 * 10^-4 = 10^-2 = 1/100. What have I missed here? Thanks Brent.


What exponent rule are you applying here?

First, √1 = 1, which means you are saying √1 * 10^-4 = 1 * 10^-4 = 10^-2, which isn't true.

Did you mean to write √10 and not √1?
Even if that's the case, the exponent property you'd need to use is: \(x^a * x^b = x^{a+b}\)

So, \(\sqrt{10} * 10^{-4} = 10^{1/2} * 10^{-4} = 10^{(1/2)+(-4)}\)
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Re: 0.0001^(1/2) [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Hi BrentGMATPrepNow, √ 1 * 10^-4 = 10^-2 = 1/100. What have I missed here? Thanks Brent.


What exponent rule are you applying here?

First, √1 = 1, which means you are saying √1 * 10^-4 = 1 * 10^-4 = 10^-2, which isn't true.

Did you mean to write √10 and not √1?
Even if that's the case, the exponent property you'd need to use is: \(x^a * x^b = x^{a+b}\)

So, \(\sqrt{10} * 10^{-4} = 10^{1/2} * 10^{-4} = 10^{(1/2)+(-4)}\)


Thanks Brent BrentGMATPrepNow for your quick reply.
Since question given is √0.00001. Therefore I'm trying to calculate it with 1 instead of 10? Is that possible?
√1 * 10^-4 = 10^-2 ( 10^-4 is under the same square root as 1 here)
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Re: 0.0001^(1/2) [#permalink]
Expert Reply
Top Contributor
Kimberly77 wrote:
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Hi BrentGMATPrepNow, √ 1 * 10^-4 = 10^-2 = 1/100. What have I missed here? Thanks Brent.


What exponent rule are you applying here?

First, √1 = 1, which means you are saying √1 * 10^-4 = 1 * 10^-4 = 10^-2, which isn't true.

Did you mean to write √10 and not √1?
Even if that's the case, the exponent property you'd need to use is: \(x^a * x^b = x^{a+b}\)

So, \(\sqrt{10} * 10^{-4} = 10^{1/2} * 10^{-4} = 10^{(1/2)+(-4)}\)


Thanks Brent BrentGMATPrepNow for your quick reply.
Since question given is √0.00001. Therefore I'm trying to calculate it with 1 instead of 10? Is that possible?
√1 * 10^-4 = 10^-2 ( 10^-4 is under the same square root as 1 here)


Using brackets will help avoid any ambiguity in the future.
One problem with your solution is that 0.00001 = 1 x 10^(-5), and not 1 x 10^(-4)

See how your calculations go with 1 x 10^(-5)
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Re: 0.0001^(1/2) [#permalink]
Thanks BrentGMATPrepNow, so error here is I'v got 1 zero less in calculation and yes it works now. You're a STAR and thanks Brent.
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Re: 0.0001^(1/2) [#permalink]
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