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please, could you explain ? And what is the level fo this question ?

Neither statement is sufficient alone; from statement 1, we might have y=1 and x=10, or y=3 and x=10, and for statement 2 we might have y=1 and x=1, or y=5 and x=1.

Combining the statements, there are a few ways one could look at this question. You have the following two inequalities, rewriting the second one so both inequalities face in the same direction:

x > 2y y + 2 > x

Recall that we can add two inequalities which face the same way, just as we add equations (be careful though - you cannot subtract inequalities in this way). Doing that here we have:

x + y + 2 > 2y + x y + 2 > 2y 2 > y

so the answer is C.

Or, perhaps more simply, you could 'chain' the inequalities together. Here we know that x > 2y, and x < y+2, so we must have that

2y < x < y + 2

So certainly 2y < y + 2, or y < 2.

Most test takers find abstract inequalities questions difficult, so any similar question would be at least a medium-high level question. One takeaway here: if you see an inequalities problem that looks like a 2 equations/2 unknowns problem, and if you don't see anything else to do, then try lining up your inequalities and adding them to see what happens. Often that will give you the answer you're looking for, as it did here. Of course there are often more direct approaches, like the one I used in the second solution above, but if you don't see that kind of solution quickly, adding your inequalities might get you there.
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Clearly each statement alone is not sufficient. When combined, you can line up both equation like \(2y<x<y+2\) (as 2y<x and x<y+2, so 2y<x<y+2), get rid of \(x\) you'll get \(2y<y+2\) --> \(y<2\), so the answer to the question is y<2 is YES. Sufficient.

Or you as the signs of the inequalities are in the opposite directions then we can subtract (2) from (1) (remember: you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(x-x>2y-(y+2)\) --> \(y<2\).

Re: If x and y are positive, is y < 2? [#permalink]

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06 Oct 2013, 09:51

Bunuel wrote:

If x and y are positive, is y< 2 ?

(1) x > 2y (2) x < y + 2

Clearly each statement alone is not sufficient. When combined, you can line up both equation like \(2y<x<y+2\) (as 2y<x and x<y+2, so 2y<x<y+2), get rid of \(x\) you'll get \(2y<y+2\) --> \(y<2\), so the answer to the question is y<2 is YES. Sufficient.

Or you as the signs of the inequalities are in the opposite directions then we can subtract (2) from (1) (remember: you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(x-x>2y-(y+2)\) --> \(y<2\).

Answer: C.

Hope it's clear.

That's great didn't know the subtraction technique. So when you subtract then the inequality sign stays the same as the minuend's in all cases right?

Clearly each statement alone is not sufficient. When combined, you can line up both equation like \(2y<x<y+2\) (as 2y<x and x<y+2, so 2y<x<y+2), get rid of \(x\) you'll get \(2y<y+2\) --> \(y<2\), so the answer to the question is y<2 is YES. Sufficient.

Or you as the signs of the inequalities are in the opposite directions then we can subtract (2) from (1) (remember: you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(x-x>2y-(y+2)\) --> \(y<2\).

Answer: C.

Hope it's clear.

That's great didn't know the subtraction technique. So when you subtract then the inequality sign stays the same as the minuend's in all cases right?

Re: If x and y are positive, is y < 2? [#permalink]

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30 Jan 2016, 10:15

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are positive. is y< 2 ?

(1) x > 2y (2) x < y + 2

When it comes to inequality, >0 can be an equation. Thus, in the original condition, there are 2 variables(x,y) and 1 equation(x,y>0), which should match with the number of equation. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), you can figure out x>2y>y, but not y<2?, which is not sufficient. For 2), in x<y+2, you cannot figure out y<2?, which is not sufficient. When 1) & 2), 2y<x<y+2, 2y<y+2, y<2 -> yes, which is sufficient. Therefore, the answer is C.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x and y are positive, is y < 2? [#permalink]

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18 Apr 2017, 12:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x and y are positive, is y < 2? [#permalink]

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01 Aug 2017, 06:53

tatane90 wrote:

If x and y are positive. is y< 2 ?

(1) x > 2y (2) x < y + 2

Given : x > 0 and y > 0 DS : y < 2

Statement 1 : x> 2y

for y = 1 , x>2. for y = 2 , x >4

NOT SUFFICIENT.

Statement 2 : x < y+2 for y = 1, x<3 for y = 2, x<4 NOT SUFFICIENT

Combined : 2y < x < y +2 For y = 1 2 <x < 3.. So there are values of x For y =2 4<x<4 No real values of x For y =3 6 <x < 5 No real values of x So for y>2, there are no real values of x.... Hence Y<2 SUFFICIENT