Are X and Y both positive ?

(1) 2x - 2y = 1
(2) x / y > 1

Statement 1: 2x-2y = 1.
2(x-y) = 1.
x-y = 1/2.
x = y + 1/2.
It's possible that y=1/2 and x=1.
It's possible that y=0 and x=1/2.
Since in the first case x and y are both positive and in the second case x and y are not both positive, insufficient.

Statement 2: x/y > 1.
It's possible that x=2 and y=1, since 2/1 > 1.
It's possible that x=-2 and y=-1, since (-2)/(-1) > 1.
Since in the first case x and y are both positive and in the second case x and y are not both positive, insufficient.

Statements 1 and 2 combined:
Statement 1: x = y + 1/2.
Statement 2: x/y > 1.
Substituting for x in the inequality:
(y + 1/2)/y > 1.
1 + 1/(2y) > 1.
1/(2y) > 0.
Thus, y>0.
Since y>0 and x = y + 1/2, we know that x>1/2.
Sufficient.

The correct answer is C.

First take-away:
The approach above combined two techniques: algebra and plugging in values.
Many DS questions are best solved using a combination of these two techniques.

Second take-away:
Given an equation with 2 variables (such as x = y + 1/2) and an inequality with the same 2 variables (such as x/y > 1), use the equation to substitute for one of the variables in the inequality.

Clearly the statements individually are insufficient.Combined"X=Y+1/2 and X/Y>1 i.e (Y+Y/2)>1 which implies Y is positive.so X also has to be positive here.Ans C
Experts is my approach correct.Or is there any faster method?

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 equations ensures a solution.


Are x and y both positive?

1) 2X - 2y = 1
2) x/Y > 1

==> In the original condition we have 2 variables (x,y) thus we need 2 equations. Since there are 1 each in 1) and 2), C is likely the answer, and it actually is.

Using both 1) & 2) together, 2x-2y=1>0 thus 2x-2y>0, 2(x-y)>0, x-y>0. In case of 2), if we multiply y^2 to both sides (the inequality sign does not change since the i multiplied positive value)) we get xy>y^2 ==> xy-y^2>0, y(x-y)>0 thus x-y>0. therefore from con 1) y>0, and x-y>0 ==> x>y>0. Having x>0 and y>0 makes the answer yes, thus it is sufficient. Therefore the answer is C.


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Hi Rich,
I have tested value for both of these before arriving at the conclusion.Anyway thanks for ur words of caution.But i am not able to get the context .Pl explain a bit more.

Thanks

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 equations ensures a solution.


Are x and y both positive?

1) 2X - 2y = 1
2) x/Y > 1
transforming the original condition and the question by variable approach method, we have 2 variables 2개(x,y). we need 2 more equations to match the number of equations and variables, and since there is 1 each in 1) and 2), C is likely the answer.

Using both 1) and 2), 2x-2y=1>0, 2x-2y>0, x-y>0, x>y. from x/y>1, multiplying both sides by y^2(since multiplying by square values maintain the direction of inequality signs) we have xy>y^2, xy-y^2>0, y(x-y)>0, therefore being x-y>0. Since y>0 and also x>y>0, meaning we have x>0, the answer is yes. The conditions are sufficient, therefore the answer is C.



If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
www.mathrevolution.com
- The one-and-only World's First Variable Approach for DS and IVY Approach for PS that allow anyone to easily solve GMAT math questions.

- The easy-to-use solutions. Math skills are totally irrelevant. Forget conventional ways of solving math questions.

- The most effective time management for GMAT math to date allowing you to solve 37 questions with 10 minutes to spare

- Hitting a score of 45 is very easy and points and 49-51 is also doable.

- Unlimited Access to over 120 free video lessons at https://www.mathrevolution.com/gmat/lesson

- Our advertising video at https://www.youtube.com/watch?v=R_Fki3_2vO8

Let's work backwards here.

S2 gives us two scenarios:

if y > 0, then x > y

or

if y < 0, then y < x

So this is close, but not quite sufficient.

S1 on its own tells us that x = y + 1/2. So x > y, but we don't know if they're both positive, both negative, or one of each.

Together, however, S1 gives us x > y, and S2 tells us that this is only possible if y > 0. So we have x > y and y > 0, forcing both variables to be positive. Sufficient!