Positive ?

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by pemdas » Fri Jan 06, 2012 3:24 pm
st(1) (x-y)=1/2 :: x and y can have any sign, Not Sufficient
st(2) implies either both are -ve or +ve along with |x|>|y|, Not Sufficient
combined st(1&2): If |x|>|y| we may have only two +ve signs for x and y according to st(2). With -ve signs of x,y it's impossible to have (x-y)=1/2 as |x|>|y|
hence x,y have +ve signs and |x|>|y|. Answer Yes, Sufficient.

c
karthikpandian19 wrote:Are x and y positive?

1. 2x-2y=1
2.x/y >1
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by karthikpandian19 » Fri Jan 06, 2012 3:35 pm
Can you explain why have you added MODULUS in your second statement and then also when combining them.
pemdas wrote:st(1) (x-y)=1/2 :: x and y can have any sign, Not Sufficient
st(2) implies either both are -ve or +ve along with |x|>|y|, Not Sufficient
combined st(1&2): If |x|>|y| we may have only two +ve signs for x and y according to st(2). With -ve signs of x,y it's impossible to have (x-y)=1/2 as |x|>|y|
hence x,y have +ve signs and |x|>|y|. Answer Yes, Sufficient.

c
karthikpandian19 wrote:Are x and y positive?

1. 2x-2y=1
2.x/y >1

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the

by pemdas » Fri Jan 06, 2012 4:17 pm
st(2) x/y >1

The numerator should be greater than the denominator and signs must be the same - both +ve or both -ve

For mathematical convenience I added modulus, as |x|>|y|.
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by Anurag@Gurome » Fri Jan 06, 2012 9:44 pm
karthikpandian19 wrote:Are x and y positive?

1. 2x-2y=1
2.x/y >1
(1) 2x - 2y = 1
Implies (x - y) = 1/2. This means the only case that is not possible is x negative and y positive. In that case the LHS will be negative always! So the possible cases are (with examples),
(1) x positive, y positive (x = 4.5, y = 4)
(2) x positive, y negative (x = 0.25, y = -0.25)
(3) x = 0.5, y = 0
(4) x = 0 , y = -0.5
(5) x negative, y negative (x = -4, y = -4.5)

(Note : In each case there is also some constraints. Like in the 1st case, not for all positive x, y the relation will hold! x must be equal to (y + 0.5) etc.) ; NOT sufficient.

(2) (x/y) > 1
Again this implies any one of the following two cases,
(1) Both of them are positive and x > y
(2) Both of them are negative and x < y; NOT sufficient.

Combining (1) and (2) together, (2) limits the number of possible cases to two. So, let's analyze them again with the help of statement 1.
(1) Both of them are positive and x > y => (x - y) = 0.5 is possible.
(2) Both of them are negative and x < y => (x - y) will be always negative. Not possible.

Thus, x and y are both positive; SUFFICIENT.

The correct answer is C.
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by sam2304 » Sat Jan 07, 2012 9:05 am
IMO C.

Are x and y +ve ?

1.x-y = 1/2. Nothing about the signs of both X and Y. INSUFF
2.x > y if both are +ve and x < y if both are -ve. INSUFF

Combining both 1 and 2.
If both are positive then x-y=1/2 is possible and hence Yes to the question.
If both are negative then x-y=1/2 cannot be arrived at. So both must be +ve.
Together SUFF.
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by MakeUrTimeCount » Mon Jan 09, 2012 12:30 am
I know everyone in the thread is supporting C. But I think it is E.

1. 2x-2y=1
=>x = y + .5
Not-sufficient either can be positive or negative.

2.x/y >1
Both should be positive or negetive.

Using both 1) and 2):
x = y + .5 and x> y

If x = -1/2 , y = -1,
x = y + .5 => True
x > y => True

If x = 1 , y = 1/2,
x = y + .5 => True
x> y => True

Answer E.

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by ronnie1985 » Mon Jan 09, 2012 1:56 am
Both the statements are not sufficient together.
First statement gives: x-y = 1/2, does not give anything about sign of x or y
Second statement just says that x>y, does not say anything about signs of x and y
Together the statements give same information that x>y by 1/2.
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