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

Question

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

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

New Question

Kurtosis · Answer

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 
eq {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

sp2018 · Answer

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

amanvermagmat · Answer

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.

US09 · Answer

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.

BhaveshGMAT · Answer

Clearly Statement 1 is insufficient.

But I am having trouble with Statement 2. Since xy > 0, we can re-write the inequality as x+y+1>y+1
Further simplifying, x>0 which means y also have to be 0 as xy>0. So, I have chosen the incorrect answer choice B.

Please guide. Where am I going wrong on this one?

bumpbot · Answer

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

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