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[48^(1/2) * 7^(1/2)]/21^(1/2) =

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Bunuel
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[48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink] New post   05 Jan 2022, 07:15
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\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} = \)

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8
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ParitoshRawat99
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Re: [48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink] New post   05 Jan 2022, 07:23
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IMO ANSWER = 4
√48.√7 = 4√3.√7
√21 = √3.√7
Cancelling √3.√7 from numerator and denominator
We get answer = 4

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BrentGMATPrepNow
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Re: [48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink] New post   05 Jan 2022, 07:24
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Bunuel wrote:
\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} = \)

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

Useful property #1: \((\sqrt{x})(\sqrt{y}) = \sqrt{xy}\)

Given: \(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} \)

Apply property #1 to get: \(\frac{\sqrt{(48)(7)}}{\sqrt{21}} \)

Find the prime factorization of numerator and denominator: \(\frac{\sqrt{(3)(2)(2)(2)(2)(7)}}{\sqrt{(3)(7)}} \)

Useful property #2: \(\frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}\)

Apply property #2 to get: \(\sqrt{\frac{(3)(2)(2)(2)(2)(7)}{(3)(7)}}\)

Simplify fraction to get: \(\sqrt{(2)(2)(2)(2)} = \sqrt{16} = 4\)

Answer: A
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Re: [48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink] New post   07 Jan 2022, 21:55
BrentGMATPrepNow wrote:
Bunuel wrote:
\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} = \)

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

Useful property #1: \((\sqrt{x})(\sqrt{y}) = \sqrt{xy}\)

Given: \(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} \)

Apply property #1 to get: \(\frac{\sqrt{(48)(7)}}{\sqrt{21}} \)

Find the prime factorization of numerator and denominator: \(\frac{\sqrt{(3)(2)(2)(2)(2)(7)}}{\sqrt{(3)(7)}} \)

Useful property #2: \(\frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}\)

Apply property #2 to get: \(\sqrt{\frac{(3)(2)(2)(2)(2)(7)}{(3)(7)}}\)

Simplify fraction to get: \(\sqrt{(2)(2)(2)(2)} = \sqrt{16} = 4\)

Answer: A


Hey BrentGMATPrepNow

I solved this way:

\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} \)

\(\frac{\sqrt{6*8} * \sqrt{7}}{\sqrt{7*3}} \)

\(\frac{\sqrt{6*8*7}}{\sqrt{7*3}} \)

\(\frac{\sqrt{3*2*4*2*7}}{\sqrt{7*3}} \)

\(\sqrt{\frac{3*2*4*2*7}{7*3}}\)

Cancel and you are left with \(\sqrt{2*4*2}\)

= 4
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BrentGMATPrepNow
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Re: [48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink] New post   08 Jan 2022, 08:06
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ritzu wrote:
BrentGMATPrepNow wrote:
Bunuel wrote:
\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} = \)

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

Useful property #1: \((\sqrt{x})(\sqrt{y}) = \sqrt{xy}\)

Given: \(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} \)

Apply property #1 to get: \(\frac{\sqrt{(48)(7)}}{\sqrt{21}} \)

Find the prime factorization of numerator and denominator: \(\frac{\sqrt{(3)(2)(2)(2)(2)(7)}}{\sqrt{(3)(7)}} \)

Useful property #2: \(\frac{\sqrt{x}}{\sqrt{y}} = \sqrt{\frac{x}{y}}\)

Apply property #2 to get: \(\sqrt{\frac{(3)(2)(2)(2)(2)(7)}{(3)(7)}}\)

Simplify fraction to get: \(\sqrt{(2)(2)(2)(2)} = \sqrt{16} = 4\)

Answer: A


Hey BrentGMATPrepNow

I solved this way:

\(\frac{\sqrt{48} * \sqrt{7}}{\sqrt{21}} \)

\(\frac{\sqrt{6*8} * \sqrt{7}}{\sqrt{7*3}} \)

\(\frac{\sqrt{6*8*7}}{\sqrt{7*3}} \)

\(\frac{\sqrt{3*2*4*2*7}}{\sqrt{7*3}} \)

\(\sqrt{\frac{3*2*4*2*7}{7*3}}\)

Cancel and you are left with \(\sqrt{2*4*2}\)

= 4

That works too!
There are several different ways to simplify that expression.
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Re: [48^(1/2) * 7^(1/2)]/21^(1/2) = [#permalink]
08 Jan 2022, 08:06
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