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

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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

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

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

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]
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.

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

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.

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If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]
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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
Subtract x and y from both sides to get: $$0 = xy - x - y$$
Great, the answer to the REPHRASED target question is YES, it IS the case that xy - x - y = 0
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 can't answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent

RELATED VIDEO FROM OUR COURSE

Originally posted by BrentGMATPrepNow on 15 Aug 2018, 06:45.
Last edited by BrentGMATPrepNow on 21 May 2022, 20:25, edited 1 time in total.
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]
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.

Hey Bunuel!

When would tautological information come into play? Statement 1, gives no new information. It just restates the information that can already be found in the prompt question.

Magoosh keeps drilling home the idea that a statement providing no new information is considered insufficient. This has burned me several times on OG questions and causes me to get a simple question wrong. Maybe I'm just not understanding the appropriate situation as to when to use the tautology idea?

I completely understand why your given explanation is the correct answer, but Magoosh rules keep replaying in my head making me want to mark Statement 1 as insufficient because of its redundancy.

Question:
Is $$(x-1)(y-1)=1$$? ----> Is $$xy - x - y + 1 = 1$$ ? ----> Is $$x + y = xy$$?

(1) $$x + y = xy$$ <---- Redundant?
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Re: If xy > 0, does (x - 1)(y - 1) = 1? [#permalink]
duchessjs 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.

Hey Bunuel!

When would tautological information come into play? Statement 1, gives no new information. It just restates the information that can already be found in the prompt question.

Magoosh keeps drilling home the idea that a statement providing no new information is considered insufficient. This has burned me several times on OG questions and causes me to get a simple question wrong. Maybe I'm just not understanding the appropriate situation as to when to use the tautology idea?

I completely understand why your given explanation is the correct answer, but Magoosh rules keep replaying in my head making me want to mark Statement 1 as insufficient because of its redundancy.

Question:
Is $$(x-1)(y-1)=1$$? ----> Is $$xy - x - y + 1 = 1$$ ? ----> Is $$x + y = xy$$?

(1) $$x + y = xy$$ <---- Redundant?

Hi duchessjs,

Remember, you need to understand that you are working not with two statements but rather with a question and a statement right?

So before discussing this question, let me provide you with an example:

Me : “John, do you want to go to the movies”

John: “Yes, I want to go to the movies”

Notice that John was able to answer the question by REPEATING what I asked him right?

So in this particular problem, the question is:

Is x + y = xy?

Statement one says: YES, x + y = xy

Thus, statement one is sufficient to answer the question.
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