Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1

Plug in approach that can be used without thinking much and very likely arrive at the correct answer.

Values to be taken: x positive and negative and find the corresponding values for y based on the statements
Note: x and y cannot be of different signs and also x cannot be zero as they will not satisfy (ii)

(i) x=10, we have y =9.5 .Both positive satisfied And now x=-10, we have y=-9.5. Both negative also satisfied .Different results. So (i) alone not sufficient

(ii) x=10, y can be positive. Both positive satisfied . And now x=-10, y can be negative. Both negative also satisfied. So (ii) alone not sufficient

(i) + (ii) x=10, y=9.5 satisfies both the statements . Both positive satisfied . And now x=-10. Value of y is found from (i) and is negative , but we see it does not satisfy (ii). So both cannot be negative .

So we can answer the question using (i) and (ii) together

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.

Hi Matthias15

If (x/y) > 1

Case-1: If y >0 then x > y

Case-2: If y <0 then x < y

Please note: Multiplying a Negative Number on both sides require the Inequation sign to flip [as mentioned in Case-2]

That's where you are wrong.

I hope it helps!

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.

Statement 1
2x-2y = 1

x - y = (1/2)

x = 3/2; y = 1 ; both x and y positive

x = 1/4, y = -1/4; x is positive, y is negative

so not sufficient

Statement 2

(x/y) > 1

x = 3/2; y = 1 ; (x/y) is more than 1. both x and y positive

x = -2; y = -1 ; (x/y) is more than 1. both x and y negative

so not sufficient

Combining Statement 1 and 2

x - y = 1/2

x = y+(1/2)

x/y > 1

{y+(1/2)} / y is more than 1

1+(1/2y) > 1

(1/2y) > 0

y has to be positive. because x = y+1/2, so x also has to be positive.

C is the answer.

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

x/0 can be really big OR really small...

From my earlier post:

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

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.

Are x and y both positive?

(1) 2x-2y = 1

2(x-y) = 1

x-y = 1/2 = 05

Lets substitute some values and check:

x = 0.1 & y = -0.4

0.1 - (-0.4) = 0.5 ==============> Answer to the question is NO

x = 1 & y = 0.5

1 - 05 = 0.5 =================> Answer to the questions is YES

As we are getting multiple answers, Statement (1) is Not Sufficient.

(2) x/y > 1

Which means both x & y have same sing, they can be either positive or both negative.

As we are getting multiple answers, statement (2) us Not Sufficient.

Now, lets combine both (1) and (2)

we get:

x = y + 1/2

also, we know x/y>1

which means: (y+1/2)/y>1

which means y is > 0

as y>0, x has to be > 0 as well. So we can come to the conclusion that both x & y are positive.


Hence, Answer is C

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!

Hi everyone,

Sorry but i dont understand the condition that 1/2y > 0 so y must be positive. But i think y can also be negative because when y is negative so 2y is still positive and 1/2y > 0 anyway. Where am i wrong? Please help. Thank you so much

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 \(\frac{1}{2y}=\frac{1}{negative}=negative<0\).

Can you please explain your statement "3rd quadrant (|x|<|y|, therefore x/y is not >1)". Why we don't need to worry about IIIrd quadrant. ?
My confusion is - Since x/y >1 - this implies both have either -ve sign or +ve sign. If both are -ve, they must lie in 3rd quadrant.

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

Hello,

1/2y > 0 means y could be 0 and 1/2y could be infinity which is greater than 0.
So we can't actually remove the point of y not equal to 0.

Regards,
Ronak B.