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If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2)

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Bunuel
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If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   05 Sep 2017, 01:19
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If x = 4 and y = 6, then \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\)


A. \(\frac{\sqrt{10}+\sqrt{15}}{10}\)

B. 1

C. \(\frac{2\sqrt{5}}{10}\)

D. \(\frac{\sqrt{10}+\sqrt{15}}{5}\)

E. 2
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D
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akshata19
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   30 Sep 2017, 09:10
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Plug in the values
√4 + √6 / √(4+6)
=2+√6 / √10

Now you see there is no option with this ans
Observe the options - fraction choices have denominator without root

Hence to remove the root multiply the numerator & denominator with √10

=(2 + √6) (√10) / (√10)(√10)
=2√10+√60 / 10
=2√10+√(4*15) /10
=2√10 + 2√15 / 10
Take common 2 out on numerator and dividing leaves denominator as 5
=√10+√15 / 5

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If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   05 Sep 2017, 03:39
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We have been asked to find the value of the expression \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}\)

Multiplying and dividing the expression by \(\sqrt{10}\) does not change the value of the expression

The expression becomes \(\frac{\sqrt{10x}+\sqrt{10y}}{\sqrt{10(x+y)}}\)
= \(\frac{\sqrt{2*5*2*2}+\sqrt{2*5*2*3}}{\sqrt{10(4+6)}}\) (Substituting x=4 and y=6)

= \(\frac{2\sqrt{5*2}+2\sqrt{5*3}}{10}\) = \(\frac{\sqrt{10}+\sqrt{15}}{5}\) (Option D)
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   30 Sep 2017, 08:47
pushpitkc wrote:
We have been asked to find the value of the expression \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}\)

Multiplying and dividing the expression by \(\sqrt{10}\) does not change the value of the expression

The expression becomes \(\frac{\sqrt{10x}+\sqrt{10y}}{\sqrt{10(x+y)}}\)
= \(\frac{\sqrt{2*5*2*2}+\sqrt{2*5*2*3}}{\sqrt{10(4+6)}}\) (Substituting x=4 and y=6)

= \(\frac{2\sqrt{5*2}+2\sqrt{5*3}}{10}\) = \(\frac{\sqrt{10}+\sqrt{15}}{5}\) (Option D)



hi

"Multiplying and dividing the expression by √10 does not change the value of the expression" that's okay ..but can you please tell me why the following method - squaring the whole expression - is not working here

(√x + √y) / √(x + y)

= { (√x + √y) / √(x + y) } ^ 2

thanks in advance
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If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   30 Sep 2017, 09:11
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Quote:
hi

"Multiplying and dividing the expression by √10 does not change the value of the expression" that's okay ..but can you please tell me why the following method - squaring the whole expression - is not working here

(√x + √y) / √(x + y)

= { (√x + √y) / √(x + y) } ^ 2

thanks in advance


Hi gmatcracker2017
it will work but the problem is the options are not in their simplest form. Hence you will have to undertake lot of calculation to arrive at the answer choice.

Straight forward Algebric method would be -
\((\sqrt{x}+\sqrt{y})/\sqrt{x+y}\), multiply the numerator and denominator by \(\sqrt{x+y}\) to get

\((\sqrt{x}+\sqrt{y})(\sqrt{x+y})/x+y\) \(=(\sqrt{x(x+y)}+\sqrt{y(x+y)})/x+y\). Now substitute the values of \(x\) & \(y\) to get

\((\sqrt{40}+\sqrt{60})/10\) \(=(\sqrt{10}+\sqrt{15})/5\)

Option D
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   30 Sep 2017, 09:35
niks18 wrote:
Quote:
hi

"Multiplying and dividing the expression by √10 does not change the value of the expression" that's okay ..but can you please tell me why the following method - squaring the whole expression - is not working here

(√x + √y) / √(x + y)

= { (√x + √y) / √(x + y) } ^ 2

thanks in advance


Hi gmatcracker2017
it will work but the problem is the options are not in their simplest form. Hence you will have to undertake lot of calculation to arrive at the answer choice.

Straight forward Algebric method would be -
\((\sqrt{x}+\sqrt{y})/\sqrt{x+y}\), multiply the numerator and denominator by \(\sqrt{x+y}\) to get

\((\sqrt{x}+\sqrt{y})(\sqrt{x+y})/x+y\) \(=(\sqrt{x(x+y)}+\sqrt{y(x+y)})/x+y\). Now substitute the values of \(x\) & \(y\) to get

\((\sqrt{40}+\sqrt{60})/10\) \(=(\sqrt{10}+\sqrt{15})/5\)

Option D


hi niks18

thanks a lot, man :-)
I got it

you are so helpful
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   29 Sep 2019, 15:24
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Solution



Approach and Working out

√x +√y/√(x+y) = √4 +√6/√(4+6)
    • = 2 +√6/√(10)
    • On rationalizing the denominator, = (2 +√6)* √(10)/√(10) * √(10)
    • = 2√10 + 2√15/10
    • = √10 + √15/5
Thus, option D is the correct answer.

Correct Answer: Option D
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AbhaGanu
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   22 Jun 2021, 00:51
How did you eliminate the 2 here? 2√10 + 2√15/10
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   22 Jun 2021, 01:11
Another way to solve this is to estimate the values of Root 4, Root 10, etc.

You basically end up with something like this for the original (4.44/3.1).

Process of elimination for the rest can be used and it's immediately clear that (B) and (E) are out.

All the best.
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink] New post   31 Aug 2023, 04:46
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Re: If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2) [#permalink]
31 Aug 2023, 04:46
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