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# root[3*root{80}+3/(9+4*root{5})]=?

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Director
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26 Jan 2012, 06:30
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74% (01:31) correct 26% (02:04) wrong based on 416 sessions

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$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$
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Re: Sqaure Root  [#permalink]

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26 Jan 2012, 06:59
5
4
LM wrote:
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$

$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3}{9+4\sqrt{5}}*\frac{9-4\sqrt{5}}{9-4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3(9-4\sqrt{5})}{81-80}}=\sqrt{12\sqrt{5}+27-12\sqrt{5}}=\sqrt{27}=3\sqrt{3}$$.

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12 Aug 2012, 09:56
Thanks Bunuel -- You Are The Man!!
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12 Aug 2012, 11:29
This can be solved by taking the denominator as 9+4\sqrt{5} and then taking this as the common factor. We get the the ans as C.
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16 Aug 2012, 23:16
This kind of problems always seemed to me very scary and requires a lot of calculation, but later i realised that GMAT never asks something that you need to calculate a lot, so one needs to look for some pattern or similar numbers/sets. In our case, we look at denominator 9+\sqrt{5} and 3\sqrt{80}, so 80 is 2^4*5, which means 4\sqrt{5}, from here we feel that numerator and denominator could be reduced. The rest is just calculations. In my opinion the most crucial part is this one.
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07 Apr 2013, 11:07
Lets analyze the first part $$3\sqrt{80} = 3\sqrt{5*16} = 3*4\sqrt{5}$$
The second term: Denominator $$(9+4\sqrt{5})*(9-4\sqrt{5})=9^2-4^2*5=1$$ Rule: (x+y)(x-y)=x^2-y^2
The second term: Numerator $$3*(9-4\sqrt{5})=27-12\sqrt{5}$$
Now putting all in one:
$$\sqrt{(}12\sqrt{5}+27-12\sqrt{5})=$$
$$\sqrt{27}=\sqrt{3*3^2}=3\sqrt{3}$$

Hope it's clear now
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02 Jan 2015, 01:12
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}$$

$$= \sqrt{3 * (4\sqrt{5}+\frac{1}{9+4*\sqrt{5}})}$$

$$= \sqrt{3 * \frac{36\sqrt{5} + 80 + 1}{9+4*\sqrt{5}}}$$

$$= \sqrt{3 * 9 * \frac{9+4*\sqrt{5}}{9+4*\sqrt{5}}}$$

$$= 3 \sqrt{3}$$
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04 Apr 2015, 07:52
2
1
LM wrote:
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$

Though, I am quite comfortable with the method mentioned by Bunuel, I found an alternative way by The Economist.

$$9 + 4*\sqrt{5}$$ will be approx. equal to 9 + 4 x 2 = 17. Hence 3/ 17 will be quite less to contribute towards the value of expression.

$$3\sqrt{80}$$ is approx. 3 x 9 = 27. now $$\sqrt{27}$$ will be something more than 5.

Now coming to options:

(A)$$\sqrt{3*\sqrt{5}}$$ is approx $$\sqrt{6}$$ which is quite less than 5. Rejected.
(B) Rejected.
(C) 3 x 1.732 = 5.1 , which is in our desired range.
(D) 3 + 4 =7. Rejected
(E) 9 + 4 x 2 = 17. Rejected.

Hence the correct ans is (C).

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27 Oct 2016, 19:41
Multiplying by the conjugate is the trick for this problem.

[3/(9+4sqrt(5))][(9-4sqrt(5)/9-4sqrt(5)] = 27-12sqrt(5)

sqrt(80) = 4sqrt(5)

Plug into equation 12sqrt(5) cancels out and we are left with sqrt(27) = 3sqrt(3)
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12 Nov 2017, 08:30
LM wrote:
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$

Let’s first simplify 3√80:

3√80 = 3 x √16 x √5 = 12√5

Next, let’s simplify 3/(9 + 4√5) by rationalizing the denominator:

3/(9 + 4√5) x (9 - 4√5)/(9 - 4√5) = (27 - 12√5)/(81 - 80) = 27 - 12√5

Finally, we have:

√(12√5 + 27 - 12√5) = √27 = √9 x √3 = 3√3

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Re: root[3*root{80}+3/(9+4*root{5})]=? &nbs [#permalink] 12 Nov 2017, 08:30
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# root[3*root{80}+3/(9+4*root{5})]=?

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