Hi,

While I see that translating (x/y)>1 into x>y must be wrong, as clearly it lets me use a different range of numbers (and is the reason I chose E instead of C), I don't understand what is wrong with my algebra.

Why can't I multiply (x/y) * y > 1 * y and get x>y?

Thank you.

What happens when y = o?

X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'.

Even in this case -

"One of the approaches:
2x−2y=1 --> x=y+12
xy>1 --> x−yy>0 --> substitute x --> 1y>0 --> y is positive, and as x=y+12, x is positive too. Sufficient."

The moment you assume y = 0, you can easily see that 1/y is infinity that is greater than 0 and hence, satisfies the equation.

Am i missing something. My answer would be 'E'

It may seem that 0.5/0 = infinity, but this is not the case.
If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER
0.5/0.1 = 5
0.5/0.01 = 50
0.5/0.001 = 500
0.5/0.0001 = 5000
0.5/0.00001 = 50000
etc.

But what if we approach 0 from the NEGATIVE side:
0.5/(-0.1) = -5
0.5/(-0.01) = -50
0.5/(-0.001) = -500
0.5/(-0.0001) = -5000
0.5/(-0.00001) = -50000
Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER

This is why we say that x/0 is undefined.

Cheers,
Brent
_________________

There are 2 variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that the correct answer is C. Using 1) and 2), from 2(x-y)=1>0 we get 2(x-y)>0, x>y. Using 2), if we multiply both sides by y^2, we get xy>y^2, xy-y^2>0, y(x-y)>0. Since x-y>0, we get y>0. From x>y>0, x>0. The answer is always yes and the condition is sufficient. Hence, the correct answer is C.



For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

_________________

What about the example of y=0? Then x/y is surely greater than 1, hence I chose answer E.

y cannot be 0 because in this case x/y will not be defined (division by 0 is not allowed) and not greater than 1 as given in the second statement.
_________________

First of all, everyone should really read Bunel's earlier posts before just tagging him. He must have written the same explanation five times!

Let me try:

Statement 1:
2x - 2y =1
2*(x-y) = 1
x - y =0.5

Ok cool, but you can probably tell intuitively that a difference of 0.5 doesn't really help. X and Y could be 1 and 0.5 or 0.25 and -0.25.

So this is not really that helpful, however it was important to do this algebra because, often on DS questions, you need to see how the statements work together and simpler expressions are much easier to work with. So, always at least try and simplify stuff wherever possible as the GMAT really seems to reward that reflex.

Statement 2:
x/y > 1

This should be an immediate reaction: same sign. If you don't immediately think that, practice until you do.

Now, as Bunel mentioned, there are two approaches. Number picking and algebra. Whenever possible, I try to use algebra. It's really personal preference on a question like this, but I have a panic that goes off in my head where I am scared I am missing a potential extreme value. So I try to use algebra to avoid that feeling.

Whenever you have an equation and an inequality, you always want to substitute the equation into the inequality (the other way around is not really permitted).

Notice that it actually doesn't matter which value you isolate. Because you know x and y have the same sign, you are just trying to figure out the sign of either one of them.

equation 1: x = y + 0.5
equation 2: x/y > 1

Now you can clean up the second equation as Bunel did, or you can go straight for it.

(y + 0.5)/y > 1
y/y + 0.5/y >1
1 + 0.5/y > 1
0.5y > 0

So a positive number (0.5) multiplied by y is greater that 0. Ok so y has to be positive. And if y is positive......so is x (because statement 2 told us that).

(C)

Keep practicing everyone!

For \(\frac{1}{2y}>0\) to be true y must be positive. It cannot be negative because in this case 2y = 2*negative = negative, so \(\frac{1}{2y}=\frac{1}{negative}=negative<0\).
_________________

Much simpler approach is here:
1) x-y=0.5
Clearly, insufficient as difference betn 2 +ve numbers as well as 2 -ve numbers can be 0.5
2) x/y>1
Insufficient, as x,y both can be either +ve or -ve

Considering both together,
x/y>1
If both are -ve, then x<y and hence x-y<0. So,there is no way that x-y=0.5 hence both cannot be zero.
If both are positive, x>y and hence x-y=0.5 can hold true.
Sufficient.

Posted from my mobile device
_________________

Dividing by 0 is not allowed. anything/0 is undefined and not infinity.
_________________