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

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Is x + y > 0 ? (I) x - y > 1 (II) x/y + 1 > 0 [#permalink]  26 Dec 2010, 07:31
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46% (02:19) correct 53% (00:09) wrong based on 13 sessions
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
[Reveal] Spoiler: OA
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Re: Inequalities [#permalink]  26 Dec 2010, 21:55
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surendar26 wrote:
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.

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Re: Inequalities [#permalink]  27 Dec 2010, 06:11
Expert's post
VeritasPrepKarishma wrote:
surendar26 wrote:
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.

Thanks. If I can , follow your every single post......are very clear
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Re: Inequalities [#permalink]  31 Jan 2011, 03:19
yeah . thanks for this too. your thinking is very clear.
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Re: Inequalities [#permalink]  31 Jan 2011, 04:46
surendar26 wrote:
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

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 yboth, at the same time, are positive or nagative...not suff.

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

Regards,
Murali.
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Check this one out - Algebra [#permalink]  04 Feb 2011, 10:56
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

(A) Statement (I) ALONE is sufficient, but statement (II) alone is not sufficient

(B) Statement (II) ALONE is sufficient, but statement (I) is not sufficient

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement alone is sufficient

(D) Each statement ALONE is sufficient

(E) Statements (I) and (II) TOGETHER are NOT sufficient
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Re: Check this one out - Algebra [#permalink]  04 Feb 2011, 11:27
Expert's post
Merging similar topics.

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

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.

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Re: Check this one out - Algebra [#permalink]  05 Feb 2011, 09:46
Bunuel wrote:
Merging similar topics.

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

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.

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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Re: Check this one out - Algebra   [#permalink] 05 Feb 2011, 09:46
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