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# If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/

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Math Expert
Joined: 02 Sep 2009
Posts: 64068
If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/  [#permalink]

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03 Sep 2017, 05:03
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:26) correct 31% (02:35) wrong based on 192 sessions

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If $$\sqrt{x} =20$$ and $$\sqrt{y} =16$$, then $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=$$

A. $$9\sqrt{41}$$

B. 2

C. $$\frac{9\sqrt{41}}{41}$$

D. 1

E. 9/41

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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/  [#permalink]

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03 Sep 2017, 06:59
1
Bunuel wrote:
If $$\sqrt{x} =20$$ and $$y = \sqrt{16}$$, then $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=$$

A. $$9\sqrt{41}$$

B. 2

C. $$\frac{9\sqrt{41}}{41}$$

D. 1

E. 9/41

Hi Bunuel,

Is there a mistake with the question? Or am i doing something wrong?
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Joined: 02 Sep 2009
Posts: 64068
Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/  [#permalink]

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03 Sep 2017, 07:04
pushpitkc wrote:
Bunuel wrote:
If $$\sqrt{x} =20$$ and $$y = \sqrt{16}$$, then $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=$$

A. $$9\sqrt{41}$$

B. 2

C. $$\frac{9\sqrt{41}}{41}$$

D. 1

E. 9/41

Hi Bunuel,

Is there a mistake with the question? Or am i doing something wrong?

_______________
Edited. Thank you.
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Math Expert
Joined: 02 Aug 2009
Posts: 8586
Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/  [#permalink]

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03 Sep 2017, 07:13
1
4
Bunuel wrote:
If $$\sqrt{x} =20$$ and $$\sqrt{y} =16$$, then $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=$$

A. $$9\sqrt{41}$$

B. 2

C. $$\frac{9\sqrt{41}}{41}$$

D. 1

E. 9/41

Hi

$$\sqrt{x} =20.....x=400$$ and $$\sqrt{y} =16,.....y=256$$,
$$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\frac{20+16}{√(400+256)}=\frac{36}{√656}=\frac{36}{4√41}=\frac{9√41}{41}$$

C
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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/  [#permalink]

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19 Sep 2018, 00:42
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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/   [#permalink] 19 Sep 2018, 00:42