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Is x^2y + xy + xy^2 > xy(1 + y)?

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Is x^2y + xy + xy^2 > xy(1 + y)? [#permalink]

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New post 25 Jan 2018, 08:07
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Is \(x^2y+xy+xy^2>xy(1+y)\)?

(1) \(y>0\)
(2) \(xy>0\)

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[Reveal] Spoiler: OA

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Re: Is x^2y + xy + xy^2 > xy(1 + y)? [#permalink]

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New post 25 Jan 2018, 09:43
\(x^2y+xy+xy^2 > xy(1+y)\)?
\(x^2y+xy+xy^2 > xy + xy^2\)?
\(x^2y > 0\)?

St1: y > 0 --> No information about x.
If x = 0 then \(x^2y = 0\)
If \(x \neq {0}\) then \(x^2y > 0\)
Not Sufficient

St2: xy > 0 --> x and y are of same sign
If x = y = -ve then \(x^2y < 0\)
If x = y = +ve then \(x^2y > 0\)
Not Sufficient

Combining both statements, y = +ve --> x = +ve. Therefore \(x^2y > 0\)
Sufficient

Answer: C
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Re: Is x^2y + xy + xy^2 > xy(1 + y)? [#permalink]

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New post 25 Jan 2018, 11:57
chetan2u wrote:
Is \(x^2y+xy+xy^2>xy(1+y)\)?

(1) \(y>0\)
(2) \(xy>0\)

New Question



question can be simplified as

xy (x + y + 1) > xy (y + 1)

so x + y + 1 > y + 1

Statement 1 says Y > 0 but nothing is mentioned about X. X can be negative or positive integer, fraction or decimal so we get both answers so not sufficient

Statement 2 says xy>0 so both x and y are positive or both of them are negative. If both positive then x+y+1 > y+1 but if both negative then x + y +1 < y +1 so not sufficient.

Combining both we get Y is positive so x is positive so given statement is true. So Answer is C
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Re: Is x^2y + xy + xy^2 > xy(1 + y)? [#permalink]

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New post 25 Jan 2018, 23:22
sajitpanicker wrote:
chetan2u wrote:
Is \(x^2y+xy+xy^2>xy(1+y)\)?

(1) \(y>0\)
(2) \(xy>0\)

New Question



question can be simplified as

xy (x + y + 1) > xy (y + 1)

so x + y + 1 > y + 1


Statement 1 says Y > 0 but nothing is mentioned about X. X can be negative or positive integer, fraction or decimal so we get both answers so not sufficient

Statement 2 says xy>0 so both x and y are positive or both of them are negative. If both positive then x+y+1 > y+1 but if both negative then x + y +1 < y +1 so not sufficient.

Combining both we get Y is positive so x is positive so given statement is true. So Answer is C


Hi Sajit

While the answer seems C to me too, I would request you to check the highlighted part. You have cancelled xy from both sides of the inequality without knowing its sign (whether positive or negative). We cannot do that in an inequality because multiplying or dividing by a negative number flips the sign of inequality. Also one of x or y might be 0, in which case we simply cannot cancel like this.

Though we can do this step in second statement and while combining the statements, because in second statement its given that xy > 0, and so by cancelling it from both sides the sign of inequality will not change.
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Re: Is x^2y + xy + xy^2 > xy(1 + y)? [#permalink]

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New post 25 Jan 2018, 23:34
IMO C

Since the statement can be rephrased as xy( x + 1 + y) > xy (1 +y)
Therefore, need to check if (x+y+1)>(y+1)
Only possible if x=+ve

1) Not sufficient, gives no idea about x

2) not sufficient, both x and y can be +ve or -ve

Combining 1) & 2), Y is +ve, therefore x will be +ve, therefore sufficient.
Re: Is x^2y + xy + xy^2 > xy(1 + y)?   [#permalink] 25 Jan 2018, 23:34
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