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surendar26
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 26 Dec 2010, 08:31
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

58% (02:06) correct 42% (01:59) wrong based on 360 sessions

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Is x + y > 0 ?

(1) x^2 - y^2 > 1
(2) x/y + 1 > 0
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E
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KarishmaB
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 26 Dec 2010, 22:55
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surendar26
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
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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.

Answer (E).
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muralimba
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 31 Jan 2011, 05:46
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surendar26
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
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Let me solve it the way GMAT expects us to do.

qtn: Is X+Y positive?

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.

Answer "E".

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Murali.
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 04 Feb 2011, 12:27
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Merging similar topics.

mariyea
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
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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.
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 28 May 2011, 07:11
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Welcome to GMAT Club!

Here is a 10-sec solution:

\(x^2, y^2, x/y\) - 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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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 29 May 2011, 15:15
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gujralvikas
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!
Show more

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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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 30 May 2018, 05:13
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VeritasPrepKarishma
surendar26
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

(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.

Answer (E).
Show more


Hi Karishma

Regarding statement 2, I understand that x and y both has to be either positive or negative. However, I was wondering if either of x or y is negative (the other being positive), we can have 2 more cases i.e if x = -1 and y = 2 then -1/2 + 1 > 0 ---> 1/2 > 0, though the answer should still be E.

Am I correct in my thinking?

Thank you.
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 30 May 2018, 05:50
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VeritasPrepKarishma
surendar26
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

(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.

Answer (E).


Hi Karishma

Regarding statement 2, I understand that x and y both has to be either positive or negative. However, I was wondering if either of x or y is negative (the other being positive), we can have 2 more cases i.e if x = -1 and y = 2 then -1/2 + 1 > 0 ---> 1/2 > 0, though the answer should still be E.

Am I correct in my thinking?

Thank you.
Show more

I have (x+y)/y > 0 in my solution. So either both (x+y) and y are positive or both are negative. The solution goes on to take those two cases.
It is certainly possible that one of x and y is positive and the other is negative (as shown by you).
The answer, of course, remains (E)
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Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 03 Mar 2025, 08:33
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