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its C

x > 2y
y+2 > x

---> x+y+2 > x+y+y
deleting X+Y both sides
2>x
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Bunuel
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?
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Bunuel
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?

Right.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.
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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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tatane90
If x and y are positive. is y< 2 ?

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

None of the statements alone is sufficient y has dependency on x, about which we do not know anything.

Multiple statement 2 by (-1) to flip the sign that makes it compatible for addition with statement 1.
x > 2y
-x>-y-2
----------
0>y-2

=> y<2

Choice C is the right option!
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