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DS / x = 3y

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Source: — Data Sufficiency |
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by real2008 » Sat Aug 29, 2009 9:38 am
is it D
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by ShikenO/kau » Sat Aug 29, 2009 9:42 am
If x = 3y, is x square2 > y square2 ?
(1) y + x < yx
(2) x square2 = 9y square2

x and y can have either have the following values:
1.Either both x and y are positive numbers
2.Or both x and y are negative numbers
3.Or both are zero

Except for the zero for all the other values of x and y x^2>y^2

Statement I:
y+x < xy only when both x and y are negative
For all negative combinations of x and y x^2is always>than y^2
So statement I is sufficient.

Statement II:
It is same as the question and so insufficient.

So my answer is A.
Please post the reply.
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by Xbond » Sun Aug 30, 2009 8:12 am
OA is A
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by pathaniaus » Sun Aug 30, 2009 12:51 pm
Statement I:
y+x < xy only when both x and y are negative
For all negative combinations of x and y x^2is always>than y^2
So statement I is sufficient.

I do not understand your method/logic. can you please explain in more detail?
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by woo » Sun Aug 30, 2009 7:48 pm
I think there is something wrong in this question because I can actually figure out the answer without any conditions.. Tell me what is wrong in my reasoning.

The question is x^2 > y^2 ?

By rearranging x^2 - y^2 > 0?

We know that x^2 - y^2 is equal to (x-y)(x+y)

thsu, (x-y)(x+y) > 0?

From the question we know that x=3y

Pluging in to the question we get

(3y-y)(3y+y)>0?

which is same as 2y*4y > 0?

8y^2 is always positive as we know hence

x^2 is always gerater than y^2.


Is something wrong?
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by ShikenO/kau » Sun Aug 30, 2009 8:43 pm
What if both x and y are 0?
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