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(25 + 25 + 100)^(1/2) =

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(25 + 25 + 100)^(1/2) =  [#permalink]

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New post 28 Oct 2018, 22:36
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

81% (00:25) correct 19% (00:26) wrong based on 70 sessions

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Re: (25 + 25 + 100)^(1/2) =  [#permalink]

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New post 28 Oct 2018, 22:42
Bunuel wrote:
\(\sqrt{25+25+100}=\)


A. \(5\sqrt{5}\)

B. \(5\sqrt{6}\)

C. \(15\)

D. \(20\)

E. \(75\sqrt{2}\)


\(\sqrt{25+25+100}\) ==> \(\sqrt{25(1+1+4)}=\)

Hence B \(5\sqrt{6}\)
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Re: (25 + 25 + 100)^(1/2) =  [#permalink]

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New post 28 Oct 2018, 22:50
√(25+25+100) = √ (5^2 (1+1+4))
= 5√6
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Re: (25 + 25 + 100)^(1/2) =  [#permalink]

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New post 28 Oct 2018, 23:03
Bunuel wrote:
\(\sqrt{25+25+100}=\)


A. \(5\sqrt{5}\)

B. \(5\sqrt{6}\)

C. \(15\)

D. \(20\)

E. \(75\sqrt{2}\)


Best way to solve this type of questions is to take out the common factor from each term to reduce calculation time.
Here we can take out \(\sqrt{25} = 5\)
Therefore, \(\sqrt{25+25+100}\)
= \(5\sqrt{1+1+4}\)
= \(5\sqrt{6}\)

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(25 + 25 + 100)^(1/2) =  [#permalink]

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New post 30 Oct 2018, 02:22
\(\sqrt{25+25+100}\)
=\(\sqrt{150}\)
=\(\sqrt{25*6}\)
=\(5\sqrt{6}\)

Option B
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Re: (25 + 25 + 100)^(1/2) =  [#permalink]

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New post 11 Nov 2018, 12:12
The GMAT may try to trick you into splitting the sum or difference of two numbers inside a radical into two individual roots. So a common mistake someone could make is to break down the numbers into its primes first, which would give you:

5 + 5 + 10 = 20.

Remember that you may only separate or combine the product or quotient of two roots. You cannot separate or combine the sum or difference of two roots. So that's why we do the following:

Calculate what is in the roots first:
√25+25+100

We get:

√150

Break it down into its primes, we get:

√3 * 2 * 5 * 5

We take the square root of the numbers:

5√6.
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Re: (25 + 25 + 100)^(1/2) = &nbs [#permalink] 11 Nov 2018, 12:12
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