Stacey and Ron---Please help me out.

[dddanny2006]

Question source-Gmat Club

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) |y - x| = x - y

I want to rephrase |y-x| please tell me if Im right.

Removing the absolute bars we get
y-x=x-y when y-x>=0 ...................................(a)
y-x=-(x-y)=y-x when y-x<=0.............................(b)

For equation a lets assume y-x=2
2=x-y ----------->x=y+2
0=x-y------------>x=y

For equation b lets assume y-x=-3
-3=y-x----------->x=y+3
y-x=0------------>x=y

Therefore statement 2 gives us 3 equations to play with.

I rephrase the main question as --

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) x=y+3 or x=y or x=y+2

Individually st 1 and st 2 are insufficient and thus we rule out option A,B and D

When we combine the St 1 and St2 we get

Pair 1- 4x=3y and x=y+3 we get y=-12 and x=-11 x>y confirmed ----YES

Pair 2- 4x=3y and x=y we get x=0 and y=0 which is not permissible due to conditions in the Question stem.


Pair 3- 4x=3y and x=y+2 we get y=-8 and x=-7 x>y

As a result C is the right answer..Is this the right way of solving this question?Is my mathematics correct when rephrasing the question?


When checking st1 individually x/y=3/4 here x<y and answer to the Question stem is No.

x/y could also be -3/-4 here x>y and the answer to the question stem is Yes.

Yes and No make statement 1 insufficient

When checking for statement 2--

x=y+3 when y is 2 ,x is 5 x>y----Yes
when y is -3, x is 0 x>y---Yes

x=y -----No


x=y+2 when y is 1,x is 3-------Yes
when y is -4 x is -2------Yes

Yes,No,Yes makes statement 2 insufficient.

My concern is that there's no overlap between statement 1 and statement 2 at all..Is that ok? I've done everything mathematically following all the rules.Im just concerned about the overlap not being there.Please clarify.












Lets assume y-x=2
Therefore 2=x-y ------> x=y+2
y-x=0--------->x=y

[RonPurewal]

I am confused. It appears that you're selecting 0, 2, and -3 as random test cases -- a good idea. But, from that point onward, you seem to be assuming that these three cases are the only possibilities in the world.

E.g., you wrote this:

I rephrase the main question as --

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) x=y+3 or x=y or x=y+2

The purple is clearly not a valid "rephrasing", because those are only 3 of the infinitude of possibilities. (What if you had tested 0, -5, and 10 instead?)
So, any conclusions you draw from this "rephrasing" are bound to be dubious.

(If you take ANY negative numbers that satisfy statement 1, you'll find that they also satisfy statement 2. Try it.)

[RonPurewal]

In any case, just remember that
|whatever| = whatever IF "whatever" is positive or 0,
|whatever| = -(whatever) IF "whatever" is negative or 0.

In the case of |y - x|, that's
= y - x IF y - x > 0,
= -(y - x) = x - y IF y - x < 0.

Because of the orange things, statement 2 just means "y - x < 0", which is the same thing as y < x.

[dddanny2006]

Hey Ron,
Well,those random values were based on conditions x>=0 and x<0

Removing the absolute bars we get
y-x=x-y when y-x>=0 ...................................(a)
y-x=-(x-y)=y-x when y-x<=0.............................(b)

I obeyed those conditions,and hence assume that they will always give you a x > y.

Is my technique wrong?

Dan

Question source-Gmat Club

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) |y - x| = x - y

I want to rephrase |y-x| please tell me if Im right.

Removing the absolute bars we get
y-x=x-y when y-x>=0 ...................................(a)
y-x=-(x-y)=y-x when y-x<=0.............................(b)

For equation a lets assume y-x=2
2=x-y ----------->x=y+2
0=x-y------------>x=y

For equation b lets assume y-x=-3
-3=y-x----------->x=y+3
y-x=0------------>x=y

Therefore statement 2 gives us 3 equations to play with.

I rephrase the main question as --

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) x=y+3 or x=y or x=y+2

Individually st 1 and st 2 are insufficient and thus we rule out option A,B and D

When we combine the St 1 and St2 we get

Pair 1- 4x=3y and x=y+3 we get y=-12 and x=-11 x>y confirmed ----YES

Pair 2- 4x=3y and x=y we get x=0 and y=0 which is not permissible due to conditions in the Question stem.


Pair 3- 4x=3y and x=y+2 we get y=-8 and x=-7 x>y

As a result C is the right answer..Is this the right way of solving this question?Is my mathematics correct when rephrasing the question?


When checking st1 individually x/y=3/4 here x<y and answer to the Question stem is No.

x/y could also be -3/-4 here x>y and the answer to the question stem is Yes.

Yes and No make statement 1 insufficient

When checking for statement 2--

x=y+3 when y is 2 ,x is 5 x>y----Yes
when y is -3, x is 0 x>y---Yes

x=y -----No


x=y+2 when y is 1,x is 3-------Yes
when y is -4 x is -2------Yes

Yes,No,Yes makes statement 2 insufficient.

My concern is that there's no overlap between statement 1 and statement 2 at all..Is that ok? I've done everything mathematically following all the rules.Im just concerned about the overlap not being there.Please clarify.












Lets assume y-x=2
Therefore 2=x-y ------> x=y+2
y-x=0--------->x=y

[dddanny2006]

Hey Ron,
Well,those random values were based on conditions x>=0 and x<0

Removing the absolute bars we get
y-x=x-y when y-x>=0 ...................................(a)
y-x=-(x-y)=y-x when y-x<=0.............................(b)


I agree that there are so many possibilities,but in all cases im sure it turns out that x>y
I obeyed those conditions,and hence tested values.And x> y.

Is my technique wrong?

Dan

In any case, just remember that
|whatever| = whatever IF "whatever" is positive or 0,
|whatever| = -(whatever) IF "whatever" is negative or 0.

In the case of |y - x|, that's
= y - x IF y - x > 0,
= -(y - x) = x - y IF y - x < 0.

Because of the orange things, statement 2 just means "y - x < 0", which is the same thing as y < x.

[RonPurewal]

If it works, it works. I wouldn't be capable of following that many things at once, but, if it works for you, go for it.

By the way, you haven't posted the official answer to this problem. That's required by the forum rules; no more discussion until the official answer is posted.
(It should be C.)

[dddanny2006]

Thanks Ron.

If it works, it works. I wouldn't be capable of following that many things at once, but, if it works for you, go for it.

By the way, you haven't posted the official answer to this problem. That's required by the forum rules; no more discussion until the official answer is posted.
(It should be C.)

[RonPurewal]

You're welcome.

Note that statement 2 is insufficient because of the possibility that x = y. For instance, if x and y are both 3, then the statement is satisfied, and that's a "no" to the question.
If x and y are not the same and statement 2 is satisfied, then, as you've already discovered, x will always be greater than y.

If you combine the two statements, then x = y is ruled out, because the only equal values that satisfy statement 1 are two 0's (which is not allowed). So, in that case, you're left with nothing but cases in which x is greater than y.