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Bunuel
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beeunoia
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Palak74
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WE:Brand Management (Retail: E-commerce)
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Palak74
I think the answer is D I and III only

From the statement we know that X>Y and from the second statement x+y<0 we know that both X and Y are negative
Thus X is a negative number greater than y

Statement I : True
Abs value x >abs value Y for our 1st question stem to hold true

Statement II : False from the main statement

Statement III : Is true since the squaring will get rid of the negative sign and a larger number will always have a higher square

What about x= -3 and y= -2, in that case second statement must be true, I haven't found a case that proves second statement false

And x/y > 1 doesn't necessarily mean x > y, what if y is negative, it doesn't hold then

I guess E is the right answer

Here's the explanation:
Since x/y > 1, we have two cases
1. x and y are positive and x> y
2. x and y are negative and x< y

From x+y< 0, we can eliminate 1 and hence we have x< y from 2.

Now as x<y and both are negative, the absolute values should have opposite relation, so |x|>|y| and same holds for the square.

So the correct option should be E

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from x/y>1 we can get: x,y both pos/neg
from x+y<0 we can determine x,y both neg
therefore from x/y>1 we can get: x<y (multiply y at both sides)
we can get: x<y<0
It's easy to tell |x|>|y|, and (x^2)>(y^2) by drawing a number axis.
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beeunoia
1. x/y > 1 => they could be both pos or neg
2. x+y<0 => x or y is neg or they both are neg

Taking into consideration 1 and 2 => both of them are negative

Let's check statements:
1. |x|>|y| => |-6|>|-2| - True
2. x<y => -6 < -2 - True
3. x^2>y^2 => (-6)^2 > (-2)^2 - True

Option E is correct!

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Why did not you take x=-2 y= -6?
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