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We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values.

Do correct me if I am wrong. _________________

The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.

I second that... C for me too.
Both x and y have to be both +ve or -ve.
Plugging in for S1 & S2 satisfies only one condition.

Good questions yezz... keep them coming.

ak_idc wrote:

I will go with C.

Here are my reasons:

We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values.

I second that... C for me too. Both x and y have to be both +ve or -ve. Plugging in for S1 & S2 satisfies only one condition.

Good questions yezz... keep them coming.

ak_idc wrote:

I will go with C.

Here are my reasons:

We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values.

Do correct me if I am wrong.

Hey guys, can't both x and y be negative too? Now what's the answer???

If we take both negative, then we can't satisfy the second condition. That means effectively we are not using the information in second condition. If we take both positive, we are fullfiling the second condition and also we can solve the problem. Hence, I say C.. _________________

The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.

Together:
From St2 we can say that either both x and y are +ve or both are -ve.

If both +ve then x>y and st1 will also be satisfied.
If both -ve then y>x then lets take out the -ve sign of both x and y. Then st1 becomes x-y = -1/2 and st2 becomes x>y and its impossible.
So both x and y are +ve. _________________

Combining both,
I says that x > y, and II says that either {x, y} are +ve or -ve
If {x, y} are -ve, then we cannot have x >y and x/y >1
Hence both x, y have to be +ve.

Picked C too. (Developing a new strategy for approaching tricky DS problems)

Making Statement 1 true by quickly picking numbers.
We can have 2(3) - 2(5/2) = 1 OR 2(-3) - 2(-7/2) = 1
x and y can be either both positive or both negative. INSUFF

From Statement 2, we only know x>y. INSUFF

Together,
If X>Y, then, in order for 2x - 2y = 1, both a and y must be positive.

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