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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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Math Expert
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Kudos [?]: 124212 [0], given: 12075

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

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05 Sep 2017, 01:19
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Question Stats:

65% (01:51) correct 35% (01:22) wrong based on 30 sessions

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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
[Reveal] Spoiler: OA

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

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05 Sep 2017, 03:39
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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Kudos [?]: 518 [0], given: 16

If x = 4 and y = 6, then (x^(1/2) + y^(1/2))/(x+y)^(1/2)   [#permalink] 05 Sep 2017, 03:39
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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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