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(I) x² - y² > 1 (x + y)(x - y) is positive. So either both are positive or both are negative. Also, absolute value of x is greater than absolute value of y. e.g. x = 3, y = 2, then (x + y) = 5 and (x+y)(x - y) = 5 x = -4, y = -2, then (x + y) = -6 and (x + y)(x - y) = 12 (x + y) can be positive or negative. Not sufficient.

(II) x/y + 1 > 0 (x+y)/y > 0 So either both are positive or both are negative. e.g. y positive. y = 4, x = 3, then (x+y) = 7 and (x + y)/y = 7/4 y negative. y = -4, x = 3, then (x+y) = -1 and (x + y)/y = (-1)/(-4) = 1/4 So x + y can be positive or negative. Not sufficient.

Taking both together, (x+y), (x -y) and y, all have the same signs. The same examples as shown for statement I above satisfy this condition. e.g. y positive. x = 3, y = 2, then (x + y) = 5, (x - y) = 1 y negative. x = -4, y = -2, then (x + y) = -6, (x - y) = -2 (x + y) can be positive or negative. Not sufficient.

stmnt1: \(x^2 - y^2 > 1\) ==> \((x+1)(x-y) > 1\) ==> \((x+y)(x-y)\) shud surely be > 0 as it is > 1 (> 1, instead of > 0, is given just to make the statement more indirect/confusing) ==> \((x+y)(x-y)\) is positive ==> \((x+y)\) and \((x-y)\) both, at the same time, are positive or nagative...not suff.

stmnt2 \(\frac{x}{y} + 1\) > 0 ==> \((x+y)/y\) > 0 ==> \((x+y)/y\) is positve again \((x+y)\) and \(y\)both, at the same time, are positive or nagative...not suff.

stmnts1 and 2 together: \(X+Y\) cab be positive or negative...NOT suff.

Simple logic would probably be the best way to deal with this question (without much calculation, algebra and number plugging).

Is x + y > 0 ?

Question asks whether the sum of \(x\) and \(y\) is positive.

(1) x² - y² > 1 --> if \(x\) is some big enough positive number and \(y\) is some small enough positive number (for example \(x=2\) and \(y=1\)) then the answer will obviously be YES as the sum of two positive values is positive BUT if you consider the same values but with the minus sign (\(x=-2\) and \(y=-1\)) then again the answer will obviously be NO as the sum of two negative values is negative. Not sufficient.

(2) x/y + 1 > 0 --> exact same approach for this statement: if both \(x\) and \(y\) are positive (which satisfies the given statement as x/y+1=positive/positive+positive) then the answer will be YES BUT if both \(x\) and \(y\) are negative (which also satisfies the given statement as x/y+1=negative/negative+positive=positive+positive) then the answer will be NO. Not sufficient.

(1)+(2) Two positive values and two negative values from (1), also satisfy (2), so we still have two answers. Not sufficient.

Simple logic would probably be the best way to deal with this question (without much calculation, algebra and number plugging).

Is x + y > 0 ?

Question asks whether the sum of \(x\) and \(y\) is positive.

(1) x² - y² > 1 --> if \(x\) is some big enough positive number and \(y\) is some small enough positive number (for example \(x=2\) and \(y=1\)) then the answer will obviously be YES as the sum of two positive values is positive BUT if you consider the same values but with the minus sign (\(x=-2\) and \(y=-1\)) then again the answer will obviously be NO as the sum of two negative values is negative. Not sufficient.

(2) x/y + 1 > 0 --> exact same approach for this statement: if both \(x\) and \(y\) are positive (which satisfies the given statement as x/y+1=positive/positive+positive) then the answer will be YES BUT if both \(x\) and \(y\) are negative (which also satisfies the given statement as x/y+1=negative/negative+positive=positive+positive) then the answer will be NO. Not sufficient.

(1)+(2) Two positive values and two negative values from (1), also satisfy (2), so we still have two answers. Not sufficient.

Answer: E.

Yeah I chose E too. I wanted to see how you would approach this q. I took the statements apart to find the q simpler to solve... Thank you Bunuel!
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Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

\(x^2, y^2, x/y\) - are insensitive to changing simultaneously signs of x and y but x+y reverses its sign. So, it's E In other words, let's say, if x=-5; y=-1 satisfies both conditions, x=5; y=1 will satisfy them too, but the answer will be different.
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I found this question in a practice test and am not able to comprehend the solution to the problem. The question is as follows:

Is x + y > 0 ? (I) x² - y² > 1 (II) x/y + 1 > 0

I marked the answer as B using the following approach, x/y + 1 > 0 x/y > -1 x > -y x + y > 0

But the correct answer this E. Can someone please explain this? Thanks!

I am not solving the question since walker has already given you a very impressive logical solution and subhashghosh has shown the solution using 'plugging in numbers'. But, let me add something here, "You multiply/divide an inequality by a number only if you know whether the number is negative or positive." If the number is positive, fine. Just go ahead and do what you do in case of equations. If the number is negative, there is no problem either but remember, you have to flip the inequality sign.

e.g. Given: x < y Multiply by 5: 5x < 5y Multiply by -5: -5x > -5y
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Re: Please help me with this question [#permalink]

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10 Jun 2014, 08:42

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Here Combining both the statements => x+y can be +ve as values can be both positive and the sum can be negative as when both will be positive so E
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Here is an interesting question I came across today. It's a good question in my opinion for all the tricks DS questions are known to present

Is x+y > 0?

(1) x^2 - y^2 > 1 (2) (x/y) + 1 > 0

Will post the OA shortly, but would be great to see some analysis here.

Cheers!

Ans: B for the first statement (1) if you plug in the numbers -4 for x and -1 for y, the statement will hold and x + y < 0 but if you plug in any two +ve numbers x + y > 0. Not Sufficient (2) (x/y) > -1 Multiply both sides by y x > -y Add y to both sides x + y > 0 Sufficient Ans: B

I would really like to see a method for the first statement and not just plug in numbers, I had to plug in 3 numbers before I could get here which took a little more than 1 min.

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