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If xy > 0, does (x - 1)(y - 1) = 1?

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If xy > 0, does (x - 1)(y - 1) = 1?

(1) x + y = xy
(2) x = y

Data Sufficiency
Question: 83
Category: Algebra First- and second-degree equations
Page: 158
Difficulty: 600


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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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SOLUTION

If xy > 0, does (x - 1)(y - 1) = 1?

\(xy>0\) means that either both \(x\) and \(y\) are positive or both are negative (so neither of unknowns equals to zero: \(x\neq{0}\) and \(y\neq{0}\)).

Question: is \((x-1)(y-1)=1\)? --> is \(xy-x-y+1=1\)? is \(x+y=xy\)?

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

(2) x = y --> question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the question becomes is \(x=2\)? We don't know that, hence this statement is not sufficient.

Answer: A.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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Bunuel wrote:

If xy > 0, does (x - 1)(y - 1) = 1?

(1) x + y = xy
(2) x = y

]


Sol: given xy> 0 means booths x and y are of same sign.let us solve the given expression we get xy -(x+y) +1

St1: x+y=xy so putting these values in the exp we get xy-xy+1 or1 so st 1 is sufficient

St2 let's put these value in the statement so we get (y-1)^2 or y^2+2y-1=0

So if y=-2 then yes expression equals 1 but if y=2 then no

So st1 is sufficient therefore ans A

600 level is okay
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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If xy > 0, does (x - 1)(y - 1) = 1?

(1) x + y = xy
(2) x = y

xy>0 means that X=0 or y=0

The question ask us if \((x - 1)(y - 1) = 1\)
Lets fix it up: \(xy-x-y-1=1\) or \(xy=x+y\)


St(1) is exacty it. Sufficient
St tells us that x=y. Not helping us much.

ans is A
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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SOLUTION

If xy > 0, does (x - 1)(y - 1) = 1?

\(xy>0\) means that either both \(x\) and \(y\) are positive or both are negative (so neither of unknowns equals to zero: \(x\neq{0}\) and \(y\neq{0}\)).

Question: is \((x-1)(y-1)=1\)? --> is \(xy-x-y+1=1\)? is \(x+y=xy\)?

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

(2) x = y --> question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the question becomes is \(x=2\)? We don't know that, hence this statement is not sufficient.

Answer: A.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 14 Jan 2015, 05:13
stm11: x+y=xy;
substitute x and y as 2 & 2 ; then the equation satisfies the stmt1 ------>>>2+2=2*2;
and also the main equation -------->>> (2-1)*(2-1) = 1;
Hence Sufficient;

Stmt2: there are many values that satisfy the equation X=Y , but do not satisfy the equation (x - 1)(y - 1) = 1.
Hence Insufficient.

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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 15 Jun 2015, 04:02
Statement one is easy enough

However statement two results in
(X)(x-2)=0 so x=0 or x=2. However x cannot be zero due to the xy>0 statement. Why is statement two not sufficient then ?

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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 15 Jun 2015, 04:06
Neethling wrote:
Statement one is easy enough

However statement two results in
(X)(x-2)=0 so x=0 or x=2. However x cannot be zero due to the xy>0 statement. Why is statement two not sufficient then ?

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How did you get x(x - 2) = 0?

Also, notice that (x - 1)(y - 1) = 1 is not given, it's what we need to find out.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 15 Jun 2015, 04:15
Yes so you got x^2=2x -> (x)(x-2) = 0 so now we need to prove this to prove the question. So when x=2 or when x=0 then (x-1)(y-1)=1 ... However we have two results and it is on this ground that it is eliminated as not sufficient. I am asking whether the statement xy>0 does not eliminate x=0 as a possible answe leaving only x=2 and this making statement two sufficient ?

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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 15 Jun 2015, 04:20
Neethling wrote:
Yes so you got x^2=2x -> (x)(x-2) = 0 so now we need to prove this to prove the question. So when x=2 or when x=0 then (x-1)(y-1)=1 ... However we have two results and it is on this ground that it is eliminated as not sufficient. I am asking whether the statement xy>0 does not eliminate x=0 as a possible answe leaving only x=2 and this making statement two sufficient ?

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(2) x = y --> question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the question becomes is \(x=2\)? We don't know that, hence this statement is not sufficient.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 15 Jun 2015, 04:36
Ah! Yes that makes more sense than the explanations above. Thanks so much.

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New post 15 Jun 2015, 04:40
Indeed. Thanks again.

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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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If xy > 0, does (x - 1)(y - 1) = 1?
One way to think of this is, are (x-1) and (y-1) reciprocals? If that is true,
y - 1 = 1/(x - 1)

(1) x + y = xy
x = xy - y => x = y (x - 1) => (x/(x-1)) = y
This can be plugged into the original equation:
( (x/(x-1) ) - 1 = (1/(x - 1) ) => multiply 1 by (x - 1) to get a common denominator => (x - (x - 1) )/ (x - 1) =(1/(x-1))
(x - x + 1)/(x - 1) = (1/(x-1))
1/(x-1) = 1/(x-1)
Sufficient.
(2) x = y
It's probably easier to try a number. For instance x = y = 8
y - 1 = 8-1 = 7
1/(x - 1) = 1/(8 - 1) = 1/7

But if x = y = 2
y - 1 = 2 - 1 = 1
1/(x - 1) = 1 (2-1) = 1
Not sufficient.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]

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New post 13 Aug 2016, 04:08
Bunuel wrote:
SOLUTION

If xy > 0, does (x - 1)(y - 1) = 1?

\(xy>0\) means that either both \(x\) and \(y\) are positive or both are negative (so neither of unknowns equals to zero: \(x\neq{0}\) and \(y\neq{0}\)).

Question: is \((x-1)(y-1)=1\)? --> is \(xy-x-y+1=1\)? is \(x+y=xy\)?

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

(2) x = y --> question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the question becomes is \(x=2\)? We don't know that, hence this statement is not sufficient.

Answer: A.



hi Bunnel.
i don't understand why statement 2 is not sufficient. From x^2 = 2x we get x=2. Why are we considering x=0 or x=2?
On substituting x=2, it satisfies the question stem.
Please help.

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New post 13 Aug 2016, 04:11
asmitap wrote:
Bunuel wrote:
SOLUTION

If xy > 0, does (x - 1)(y - 1) = 1?

\(xy>0\) means that either both \(x\) and \(y\) are positive or both are negative (so neither of unknowns equals to zero: \(x\neq{0}\) and \(y\neq{0}\)).

Question: is \((x-1)(y-1)=1\)? --> is \(xy-x-y+1=1\)? is \(x+y=xy\)?

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

(2) x = y --> question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the question becomes is \(x=2\)? We don't know that, hence this statement is not sufficient.

Answer: A.



hi Bunnel.
i don't understand why statement 2 is not sufficient. From x^2 = 2x we get x=2. Why are we considering x=0 or x=2?
On substituting x=2, it satisfies the question stem.
Please help.


Two solutions satisfy x^2 = 2x, x=0 and x=2.
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Re: If xy > 0, does (x - 1)(y - 1) = 1?   [#permalink] 01 Sep 2017, 08:13
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