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

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

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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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New post 09 Feb 2014, 23:30
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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 10 Feb 2014, 10:00
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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

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

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New post 10 Feb 2014, 15:49
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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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New post 18 May 2015, 02:49
Bunnel,

Option B turns out to be x=0 or x=2.X cannot be equal to zero but we can use x=2 right.In that case we will get (x-1)(y-1)=1
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Re: If xy > 0 does (x-1)(y-1)=1 ?  [#permalink]

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New post 18 May 2015, 03:29
kirtivardhan wrote:
Bunnel,

Option B turns out to be x=0 or x=2.X cannot be equal to zero but we can use x=2 right.In that case we will get (x-1)(y-1)=1


Dear kirtivardhan

Your question tells me that you analyzed St. 2 in one of the following two ways. I'll list them both here and discuss the error in them.

Way 1- You did your analysis of St. 2 as follows:

"Put x = y in x + y = xy
=> \(2x = x^2\)
Upon solving, x = 0 or x = 2"


The error a student who analyses St. 2 in this way does is that he is not considering the equation x = y alone (which is the only piece of info that St. 2 gives) but instead has mistakenly carried over information from St. 1 (x + y = xy) into his analysis of St. 2.

Way 2- You did your analysis of St. 2 as follows:

"Given that x = y
We need to find if \((x-1)^2 = 1\)?
That is, if \((x-1)^2 - 1 = 0\) or (x-1-1)(x-1+1) = 0
That is, x(x-2) = 0
That is, x = 0 or x = 2

We're given that xy > 0. This means, x cannot be EQUAL TO zero. So, x = 2"


The error a student who analyses St. 2 in this way does is that he has used the equation \((x-1)^2 = 1\) as a FACT, not as something to be verified

____________


The correct analysis of St. 2 would be as under:

From the question statement, we know that x and y have same sign and both are not equal to zero

From St. 2, x = y

Using this, we've to determine if (x-1)(y-1) = 1? That is, if \((x-1)^2 = 1\)? That is, if x(x-2) = 0

That is, we need to determine if x = 0 or x = 2.

We know that x cannot be equal to 0

So, we need to determine if x = 2. If x = 2, the equation in the question will hold true. For other values of x, the equation in the question will not hold true.

Since we don't know if x = 2 or not, St. 2 is insufficient to arrive at a unique answer for the given question.


Hope this discussion helped! :)

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

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New post 18 May 2015, 03:39
Thanks Japinder,

You mean to say that we plugged option b in the question asked and we got x=0 and x=2.If x would have been mentioned to be 2 somehow then option b would have been suff.

I am making sense?

In short ,what we are trying to do is we are plugging an option in to the question asked and checking whether the result is there in the fact

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

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New post 18 May 2015, 04:05
kirtivardhan wrote:
Thanks Japinder,

You mean to say that we plugged option b in the question asked and we got x=0 and x=2.If x would have been mentioned to be 2 somehow then option b would have been suff.

I am making sense?

In short ,what we are trying to do is we are plugging an option in to the question asked and checking whether the result is there in the fact

Regards


Dear kirtivardhan

Yes, you're absolutely right in your first statement. If we were given, either in the question statement or in St. 2 itself that x = 2, then Option B would have been sufficient.

To answer your last statement, let me reiterate what we are trying to do here in Analysis of St. 2:

1. Info given in Question statement: xy > 0. This is a FACT

2. Info given in St. 2: x = y. This is also a FACT

Using these 2 facts, we need to confirm if (x-1)(y-1) = 1? This is the QUESTION.

By using Facts 1 and 2 to simplify the question, we saw that the answer to this question is YES, if x = 2 and NO, if x has some other value.

Hope this clarification helped! :)

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

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New post 19 May 2015, 02:57
Bunuel wrote:
guytree wrote:
I am bit sceptic to post this. But I wanted to check if this approach is right.

From the question we know that X and Y both are greater than 0.

In the statement 2 we could use simple plug-ins. If x=y=2 then (x-1)(y-1)=1. However, if x=y=3 then (x-1)(y-1) is not equal to 1.

I would greatly appreciate if you let me understand any loopholes in this approach.

Cheers


If \(xy>0\) does \((x-1)(y-1)=1\)?
(1) \(x + y = xy\)
(2) \(x=y\)

\(xy>0\) 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 us the answer YES. 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 Bunuel,

I cannot find any values that represent xy = x+y
Can you provide some explanation?
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Re: If xy > 0 does (x-1)(y-1)=1 ?  [#permalink]

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New post 19 May 2015, 03:12
LaxAvenger wrote:
Bunuel wrote:
If \(xy>0\) does \((x-1)(y-1)=1\)?
(1) \(x + y = xy\)
(2) \(x=y\)

\(xy>0\) 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 us the answer YES. 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 Bunuel,

I cannot find any values that represent xy = x+y
Can you provide some explanation?


LaxAvenger
xy = x+y

So, xy - x = y
=> x(y-1) = y
=> \(x = \frac{y}{(y-1)}\)

From this equation, you can find a number of (x,y) pairs.

Example, when y = 2, x = 2
When y = 3, x = 3/2 etc.

Please note that you're not told that x and y are integers. So, you should not assume it.

Hope this helped! :)

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

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New post 16 Jun 2015, 09:18
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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 21 May 2017, 04:17
jakolik wrote:
Hi,

The question is:
(x-1)(y-1)=1 or xy-y-x+1=1 or xy=y+x

Thus first statement is sufficient.
Second statement x=y
x^2=2x which is not sufficient to answer the question.

So the right answer should be A. Are you sure the OA is C?

regards,
Jack



hey just a question, why can we not divide both sides by x to get x=2

I know its wrong but could someone pls let me know why it is wrong ?
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Re: If xy > 0 does (x-1)(y-1)=1 ?  [#permalink]

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New post 23 May 2017, 05:51
daviddaviddavid wrote:
jakolik wrote:
Hi,

The question is:
(x-1)(y-1)=1 or xy-y-x+1=1 or xy=y+x

Thus first statement is sufficient.
Second statement x=y
x^2=2x which is not sufficient to answer the question.

So the right answer should be A. Are you sure the OA is C?

regards,
Jack



hey just a question, why can we not divide both sides by x to get x=2

I know its wrong but could someone pls let me know why it is wrong ?



Hey,

If we do not whether the variable x is greater than zero or not, we cannot divide both sides by x and get x = 2.

    \(x^2 = 2x\)
    \(x^2 -2x = 0\)
    \(x(x-2) = 0\)
Therefore the value of x can be 0 or 2.

If you divide \(x^2 = 2x\) by \(x\) on both sides, you are assuming that x is not equal to 0 and thus this division would make sense: \(\frac{x^2}{x} = \frac{2x}{x}\) .

If \(x = 0\) then, \(\frac{x^2}{0} = \frac{2x}{0}\) is not valid as such. Hence, if one does not know whether x is 0 or non-zero, we take all the terms to the left-hand side of the equation and then solve for the value of x.

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

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New post 20 Feb 2018, 16:18
Bunuel wrote:

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

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


We can re-express the question as:

Does xy - x - y + 1 = 1 ?

Does xy = x + y ?

Statement One Alone:

x + y = xy

We see that statement one answers the question.

Statement Two Alone:

x = y

Knowing that x = y, is not sufficient to answer the question. If x = y = 2, then (2 - 1)(2 - 1) = 1; however, if x = y = 1, then (1 - 1)(1 - 1) does not does equal 1.

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

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New post 15 Aug 2018, 05:45
Top Contributor
Bunuel wrote:
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

Given: xy > 0

Target question: Does (x - 1)(y - 1) = 1?

This is a good candidate for rephrasing the target question
Take the equation: (x - 1)(y - 1) = 1
Use FOIL to expand the left side to get: xy - x - y + 1 = 1
Subtract 1 from both sides to get: xy - x - y = 0

REPHRASED target question: Does xy - x - y = 0?

Statement 1: x + y = xy
Take the REPHRASED target question and replace xy with x+y to get: Does (x + y) - x - y = 0?
Simplify the left side to get: Does 0 = 0?
Great! The answer to the REPHRASED target question is YES, it IS the case that xy - x - y = 0
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = y
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 2 and y = 2. In this case, the equation xy - x - y = 0 becomes (2)(2) - 2 - 2 = 0, which works!
So, the answer to the REPHRASED target question is YES, it IS the case that xy - x - y = 0

Case b: x = 1 and y = 1. In this case, the equation xy - x - y = 0 becomes (1)(1) - 1 - 1 = 0, which does NOT work.
So, the answer to the REPHRASED target question is NO, it is NOT the case that xy - x - y = 0

Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

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Re: If xy > 0, does (x - 1)(y - 1) = 1? &nbs [#permalink] 15 Aug 2018, 05:45
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