Question:
Is xy + xy < xy ? (1) x^2 / y < 0 (2) x^9 (y^3)3 < (x^2)4 y^8
My Solution:
Rephrase the Question: is xy<0? or are x and y of opposite sign?
From 1. Y is negative. No idea about X - Not sufficient
From 2. (XY)^9 - (XY)^8 < 0
(XY)^8 (XY - 1)<0
(XY)^8 - always positive, therefore XY-1<0 or XY<1 - again XY = 0 or XY<0 - Not sufficient
1 + 2
y<0 and XY<1
now x can be equal to 0 (So the Q: xy<0 will be a NO)
OR x can be a negative number (Again: Q: xy<0 will be a NO since x is negative and y is negative which will give a positive xy and xy<0 won't be possible)
OR x can be a positive number (So the Q: xy<0 will be a YES)
Therefore I selected the answer as E
E is the right answer but I didn't get much when I tried to look at the solution and understand the different approach and therefore would need help to understand whether the way in which I arrived at the answer is correct or wrong.
Please help.
Thanks
Is xy + xy < xy ? (1) x^2 / y < 0 (2) x^9 (y^3)3 < (x^2)4 y^8
My Solution:
Rephrase the Question: is xy<0? or are x and y of opposite sign?
From 1. Y is negative. No idea about X - Not sufficient
From 2. (XY)^9 - (XY)^8 < 0
(XY)^8 (XY - 1)<0
(XY)^8 - always positive, therefore XY-1<0 or XY<1 - again XY = 0 or XY<0 - Not sufficient
1 + 2
y<0 and XY<1
now x can be equal to 0 (So the Q: xy<0 will be a NO)
OR x can be a negative number (Again: Q: xy<0 will be a NO since x is negative and y is negative which will give a positive xy and xy<0 won't be possible)
OR x can be a positive number (So the Q: xy<0 will be a YES)
Therefore I selected the answer as E
E is the right answer but I didn't get much when I tried to look at the solution and understand the different approach and therefore would need help to understand whether the way in which I arrived at the answer is correct or wrong.
Please help.
Thanks
