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by gmatusa2010 » Sun Dec 05, 2010 10:58 pm
If xy > o, does (x-1)(y-1) = 1?

1. x+y = xy

2. x = y

Is there a way to do this without expanding (x-1)(y-1)?

I didn't immediately see that x+y=xy is the same thing. I was actually distracted by xy>0. Is there a way to use xy>0?
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Source: — Data Sufficiency |
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by shovan85 » Sun Dec 05, 2010 11:16 pm
gmatusa2010 wrote:If xy > o, does (x-1)(y-1) = 1?

1. x+y = xy

2. x = y

Is there a way to do this without expanding (x-1)(y-1)?

I didn't immediately see that x+y=xy is the same thing. I was actually distracted by xy>0. Is there a way to use xy>0?
1. x+y = xy

This will happen only when x = y = 0 OR x = y = 2

The reason xy>0 is given so that you can not select x = y = 0

Thus sufficient.

2. x = y

Insufficient as x and y can be any number.

IMO A
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by goyalsau » Sun Dec 05, 2010 11:23 pm
gmatusa2010 wrote:If xy > o, does (x-1)(y-1) = 1?

1. x+y = xy

2. x = y

Is there a way to do this without expanding (x-1)(y-1)?

I didn't immediately see that x+y=xy is the same thing. I was actually distracted by xy>0. Is there a way to use xy>0?
xy > 0, Means Whether x and y are both positive or Both Negative.

1 ). x + y =xy, By This we can see that x and y has to be positive,

Because when we add two negative result in always negative , but when we multiply two negative result is always positive.

so By This we can discard all the negative values for x and y .

x and y has to be equal to 2,.

Because 2 is the only integer those square is equal to the addition its own.

take for example x = 3 and y = 1 x +y = xy { 3 + 1 = 3 } Not possible
Take fractions When we multiply two fractions result will always be a smaller fraction , On the other hand when we add two positive fractions result is always a value higher than any individual value of the fraction.

1/2 + 1/2 = 1, But 1/2 * 1/2 = 1/4 ,

so 2 is the only possible integer,

So its sufficient

2 ) x = y
there is no need for this equation as we all ready know that the value of x is equal to y,

so answer is A
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by Rahul@gurome » Mon Dec 06, 2010 2:25 am
gmatusa2010 wrote:If xy > o, does (x-1)(y-1) = 1?

1. x+y = xy

2. x = y

Is there a way to do this without expanding (x-1)(y-1)?

I didn't immediately see that x+y=xy is the same thing. I was actually distracted by xy>0. Is there a way to use xy>0?
In this question, statement 1 is more than sufficient. If we have statement 1 available, we don't need any extra information to say that (x - 1)(y - 1) = 1. Because,
.... x + y = xy
=> xy - x - y = 0
=> xy - x - y + 1 = 1
=> (x - 1)(y - 1) = 1

Thus if (x + y) = xy, it implies that (x - 1)(y - 1) is always equal to 1. Even for x = 0, y = 0. As the question doesn't ask for the possible values of x and y, the information xy > 0 is useless for statement 1.
shovan85 wrote:1. x+y = xy
This will happen only when x = y = 0 OR x = y = 2
The reason xy>0 is given so that you can not select x = y = 0
Nowhere it is mentioned x and y are integers.
Take x = 3 and y = 3/2 => (x + y) = xy
Take x = 4 and y = 4/3 => (x + y) = xy
Take x = 10 and y = 10/9 => (x + y) = xy
In fact if (x + y) = xy, then rearranging terms results y = x/(x - 1). For any value of x (except x = 1), there exists a value y for which statement 2 satisfies.

And another point the question doesn't ask for possible value of x and y. Thus there is no point whether we can select x = y = 0 or not.
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by shovan85 » Mon Dec 06, 2010 3:28 am
Rahul@gurome wrote:
shovan85 wrote:1. x+y = xy
This will happen only when x = y = 0 OR x = y = 2
The reason xy>0 is given so that you can not select x = y = 0
Nowhere it is mentioned x and y are integers.
Take x = 3 and y = 3/2 => (x + y) = xy
Take x = 4 and y = 4/3 => (x + y) = xy
Take x = 10 and y = 10/9 => (x + y) = xy
In fact if (x + y) = xy, then rearranging terms results y = x/(x - 1). For any value of x (except x = 1), there exists a value y for which statement 2 satisfies.

And another point the question doesn't ask for possible value of x and y. Thus there is no point whether we can select x = y = 0 or not.
Yes!! Rahul True Very True....

Thanks a lot :)
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by diebeatsthegmat » Mon Dec 06, 2010 10:33 am
gmatusa2010 wrote:If xy > o, does (x-1)(y-1) = 1?

1. x+y = xy

2. x = y

Is there a way to do this without expanding (x-1)(y-1)?

I didn't immediately see that x+y=xy is the same thing. I was actually distracted by xy>0. Is there a way to use xy>0?
or you can solve it by the following way. it took me only 25 second to solve it by this way
(x-1))y-1)=xy- (x+y)+1=1? <=> xy=x+y
see! A is sufficient and B is not
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