DS Question---Is the technique correct?No overlap found :(

[dddanny2006]

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.

[MiKeKiM]

Hi there --

Not sure if what you wrote represents your real-time thought process, but whenever you see absolute value on the GMAT, especially in Data Sufficiency, I think it's a good idea to think about it conceptually before doing any computation-that is, try to figure out the significance of what the absolute value information is telling you before you break it down into equations and start doing nitty-gritty math work.

Obviously there are different ways to go about solving a problem like this, but just to illustrate, this is what my real-time thoughts were looking at the q for the first time --

Statement one: x = (3/4) y. Means x is smaller if they are both positive, x is greater if they are both negative. Can't answer.
Statement two: x - y must be positive, or zero. If x - y is positive, that must mean x is bigger than y. If x - y =0, x and y are equal. Can't answer w/this alone but...

Now I know if I combine both statements, x and y are not equal, so x - y must be positive, and x must be bigger than y.

Hope that helps -- Mike Kim

[theCodeToGMAT]

To find: x > y

Statement 1:
4x = 3y
x = 3/4 * y
If y is positive, x < y
If y is negative, x > y
INSUFFICIENT

Statement 2:
|y - x| = x - y
We know that, |y - x| will be either 0 or positive
So, x - y >= 0
x - y > 0 when x != y
x - y = 0 when x = y
we cannot say whether x>y
INSUFFICIENT

Combining..
x != y
So,
x - y > 0
That implies x > y
SUFFICIENT

Answer [spoiler]{C}[/spoiler]

[theCodeToGMAT]

dddanny2006, you complicated the solution too much..

Please refer my solution and let me know if you have any difficulty in understanding that.

[dddanny2006]

Hi Rahul

Yes I did complicate the procdeure because I wanted to make sure I got the concept of removing the absolute bars right subjecting to conditions x>=0 and x<0.
Yet,tell me if my methodology is correct.Is the math right?

I like your solution,which book did you refer?Manhattan doesnt teach it that ways.

Thanks again

Dan

dddanny2006, you complicated the solution too much..

Please refer my solution and let me know if you have any difficulty in understanding that.

[Mathsbuddy]

Ignoring any cases of x=y=0 gives

(1)4x = 3y which could mean x>y or x<y, so ignore it as it's insufficient.

(2)|y - x| = x - y this is a clear statement that x > y

because modulus is always >= 0, so x >= 0, but we need to ignore x = 0


Therefore x > y. True.