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If xy # 0, is (1/x + 1/y) = 2 ?

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If xy # 0, is (1/x + 1/y) = 2 ? [#permalink]

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New post 20 Sep 2010, 11:57
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If xy # 0, is (1/x + 1/y) = 2 ?

(1) x = y
(2) x + y = 2xy
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Re: Algebra DS [#permalink]

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tatane90 wrote:
if xy != 0, is (1/x + 1/y) = 2 ?
1) x=y
2) x+y = 2xy

Any idea ?

Thanks


Is \(\frac{1}{x}+ \frac{1}{y}= 2\)? --> is \(\frac{x+y}{xy}=2\)? --> is \(x+y=2xy\)?

(1) \(x=y\) --> question becomes is \(2x=2x^2\)? --> reduce by \(2x\) (we can safely do this as \(x\neq{0}\)) --> is \(x=1\)? We don't know that. Not sufficient.

(2) \(x+y = 2xy\) --> directly gives us the YES answer to the question. Sufficient.

Answer: B.
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Re: Algebra DS [#permalink]

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New post 20 Sep 2010, 13:38
tatane90 wrote:
if xy != 0, is (1/x + 1/y) = 2 ?
1) x=y
2) x+y = 2xy

Any idea ?

Thanks


1. say x=y=2 xy=4 1/2+1/2 != 2
BUT x=y=1 xy=1 1/1 +1/1 = 2! so INSUFF

2.
you could see it another way - if x and y are the same then their sum equals to 2x or 2y and that cant be 2xy. the only way to be 2xy is if one of them is 1. so..
1 +y = 2(1)y = 1+y =2y. so y has to be 1 to satisfy 2y there fore 1/1 +1/1 =2 so SUFF
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Re: Algebra DS [#permalink]

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New post 20 Sep 2010, 18:20
tatane90 wrote:
if xy != 0, is (1/x + 1/y) = 2 ?
1) x=y
2) x+y = 2xy

Any idea ?

Thanks


1. (1/x + 1/y)
= x+y/xy
=2y/y^2
=2/y
Insufficient.

2. x+y/xy
=2xy/xy
=2

Hence the answer is B.
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Re: Algebra DS [#permalink]

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New post 20 Sep 2010, 18:53
tatane90 wrote:
if xy != 0, is (1/x + 1/y) = 2 ?
1) x=y
2) x+y = 2xy
Thanks


(1) x = y

1/x + 1/y = 1/x + 1/x = 2/x can't tell if 2/x = 2
1/x + 1/y = 1/y + 1/y = 2/y can't tell if 2/y = 2
insufficient

(2) x + y = 2xy

Divide both sides by x:
>> 1 + y/x = 2y
Divide both sides by y:
>> 1/y + 1/x = 2
sufficient

Ans B
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Re: If xy # 0, is (1/x + 1/y) = 2 ? [#permalink]

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Re: If xy # 0, is (1/x + 1/y) = 2 ?   [#permalink] 12 Apr 2016, 07:39
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