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

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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.
Source: — Data Sufficiency |

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by MiKeKiM » Sat Nov 16, 2013 10:08 pm
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

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by theCodeToGMAT » Sun Nov 17, 2013 12:59 am
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]
R A H U L

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by theCodeToGMAT » Sun Nov 17, 2013 1:35 am
dddanny2006, you complicated the solution too much.. :)

Please refer my solution and let me know if you have any difficulty in understanding that.
R A H U L

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by dddanny2006 » Sun Nov 17, 2013 5:12 am
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
theCodeToGMAT wrote:dddanny2006, you complicated the solution too much.. :)

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

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by Mathsbuddy » Sun Nov 17, 2013 12:18 pm
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.