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Bunuel
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BrentGMATPrepNow
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Somnath09
Why not B. 3/sqrt 3 is in B

Sent from my XT1068 using GMAT Club Forum mobile app

Here's another way to think of it.
We know that (sqrt 3)(sqrt 3) = 3, so let's replace the numerator, 3, with (sqrt 3)(sqrt 3)

We get: 3/sqrt 3 = (sqrt 3)(sqrt 3)/(sqrt 3) =sqrt 3

Cheers,
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Taking \(\sqrt{45}\) common from numerator and denominator
\(\sqrt{45}\)(\(\sqrt{4}\)+1) / \(\sqrt{45}\)*(\(\sqrt{3}\))
(2+1)/\(\sqrt{3}\)
3/\(\sqrt{3}\)
\(\sqrt{3}\)
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Bunuel
\(\frac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

A. \(\sqrt{3}\)

B. \(3\sqrt{3}\)

C. \(5\sqrt{3}\)

D. \(3\sqrt{5}\)

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

We can start by simplifying √180, √45, and √135:

√180 = √36 x √5 = 6√5

√45 = √9 x √5 = 3√5

Thus, 6√5 + 3√5 = 9√5.

√135 = √9 x √15 = 3√15

We now have:

(9√5)/(3√15) = 3/√3

Finally, we rationalize the denominator, getting:

(3/√3)(√3/√3) = 3√3/3 = √3

Answer: A
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(√180 - √45)/√135 = (6√5 - 3√5)/3√15
= 9√5 / 3√15
= 3 / √3
= √3
Answer A
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