AB78,

For example:

For example:

Most of the people forget that an equation such as X^2=2X has 2 solutions!

X^2 - 2X = 0

x( X- 2) = 0

SO X = 0 and X = 2 ( People usually forget about X )

Going back to our problem:

(x-1)(y-1) = 1 can be simplified as X+y=xy as in Statement (1)

if you try to find X from this equation:

xy-x=y

x(y-1)=y

X = Y / Y-1

So first of all, you can see that Y can't be equal to 1! so x=y=1 is not s solution for this equation.

But x=y=2 is

INSUF

so it is A.

Hope it helps!

I would say D is the answer. Please post if you find out why B would be insufficient.

Thanks.

ab78 wrote:If xy > 0 , does (x-1)(y-1) = 1 ?

1. x+y = xy
2. x=y

OA = A

I got it as D and I can explain why. Don't know if I am doing some stupid mistake.

1 . Is sufficient pretty straightforward.
2. (y-1)(y-1)=1 thus y^2 - 2y=0 thus we get either y=0 or y=2. since they have mentioned xy>0 y can't be zero, it has to be 2 and hence x=2 and I am getting Sufficient answer. If the OA in OG is correct then I am really doing some real stupid mistake.

Can someone point out what mistake I am doing?
[/spoiler]

Picking numbers clearly illustrates that (2) isn't sufficient.

Does (x-1)(y-1) = 1?

(2) x = y

if we pick x=y=2, then we get the question:

Does 1*1 = 1? YES

if we pick x = y = 3, then we get the question:

Does 2*2 = 1? NO

The mistake you made was that you didn't try to answer the question - you assumed that (x-1)(y-1) = 1 is a fact, which is not the case.

Remember, the STATEMENTS are facts that we take to be true. The QUESTION is what we're trying to figure out.

So, when dealing with (2) algebraically, we get:

Is (y-1)(y-1) = 1?

Is y^2 - 2y - 1 = 1?

Is y^2 - 2y = 0?

Is y(y-2) = 0?

Well, if y=0 or y=2, then YES it is (and as you pointed out, y can't = 0, but it still could = 2).

However, if y is any other number, then NO it isn't.

So, the answer to the question could be YES or NO: insufficient.

rock1981 wrote:I would say D is the answer. Please post if you find out why B would be insufficient.

Thanks.

(x-1)(y-1) = 1 ?

What happens when you have X=Y=1

Will you get 1 ? or 0 ?

Thanks everyone for the reply. I appreciate the responses and I now know where I went wrong.

Logitech on different note your signature is pretty interesting

A)

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

1. x+y = xy


Simplify (x-1)(y-1) = 1
xy-x-y+1=1
xy-x-y=0

Stmt I) is exactly this

xy-x-y=0 No proof necessary

Awesome Answer and a such a a painfully obvious thing pointed out.

Great. Thanks.

ab78 wrote:If xy > 0 , does (x-1)(y-1) = 1 ?

1. x+y = xy
2. x=y

OA = A

I got it as D and I can explain why. Don't know if I am doing some stupid mistake.

1 . Is sufficient pretty straightforward.
2. (y-1)(y-1)=1 thus y^2 - 2y=0 thus we get either y=0 or y=2. since they have mentioned xy>0 y can't be zero, it has to be 2 and hence x=2 and I am getting Sufficient answer. If the OA in OG is correct then I am really doing some real stupid mistake.

Can someone point out what mistake I am doing?
[/spoiler]

A very simple approach!!

given xy >0
so neither x nor y would be 0

1) x + y = xy

x( y-1) = y

since x can't be 0
so y-1 = y/x

on same lines x-1 = x/y

hence (x-1)*(y-1) = 1

suff..


2) x = y

(x-1)^2 ?

we don't know the value of x

hence

insuff..


Logitech,

X = Y / Y-1

So first of all, you can see that Y can't be equal to 1! so x=y=1 is not s solution for this equation. ????