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# If xy ≠ 0, is 1/x +1/y = 6?

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Joined: 02 Sep 2009
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If xy ≠ 0, is 1/x +1/y = 6?  [#permalink]

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24 Dec 2017, 01:52
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45% (medium)

Question Stats:

71% (01:09) correct 29% (01:49) wrong based on 38 sessions

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If xy ≠ 0, is 1/x +1/y = 6?

(1) x = y

(2) x + y = 2/3

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Re: If xy ≠ 0, is 1/x +1/y = 6?  [#permalink]

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24 Dec 2017, 03:10
1
Bunuel wrote:
If xy ≠ 0, is 1/x +1/y = 6?

(1) x = y

(2) x + y = 2/3

Awaiting OA

Statement 1: insufficient; x & y can be 1/3 that makes up 6. x & y can be 1 which makes up 2

Statement 2: insufficient; there can be multiple values possible for such a compbination

Statement 1 & 2: x=y=1/3; sufficient
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Re: If xy ≠ 0, is 1/x +1/y = 6?  [#permalink]

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24 Dec 2017, 03:39
IMO:C
St:1 gives 2/x. x can be anything. Insufficient.
St:2 gives x+y. We cannot deduce i/x+i/y from that. Insufficient.
St1+St2
x+x=2/3
=>x=1/3.
We can calculate y and thus can calculate 1/x+1/y.
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Re: If xy ≠ 0, is 1/x +1/y = 6?  [#permalink]

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24 Dec 2017, 14:18
1
Bunuel wrote:
If xy ≠ 0, is 1/x +1/y = 6?

(1) x = y

(2) x + y = 2/3

So x and y are not 0, but they could be either a negative or a positive.

Given: 1/x+1/y = x+y/xy.

(1) x = y. $$x+y/xy = 2x/x^2 = 2/x$$, we don't know the values for either, so insufficient.

(2) x + y = 2/3. $$x+y/xy = (2/3)/xy.$$, we don't know the values for either, so insufficient.

(1) & (2) $$x+y = x+x = 2/3$$, so $$2x=2/3$$ and $$x=(2/3)/2=1/3$$; now, $$x+y/xy = 2x/x^2 = 2/x = 2/(1/3) = 6$$, sufficient.

Re: If xy ≠ 0, is 1/x +1/y = 6? &nbs [#permalink] 24 Dec 2017, 14:18
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