DS / x = 3y
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Source: Beat The GMAT — Data Sufficiency |
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ShikenO/kau
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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.
(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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pathaniaus
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- Joined: Wed May 06, 2009 8:57 am
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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?
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?
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?
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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ShikenO/kau
- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Sat Jul 11, 2009 8:37 pm
- Thanked: 2 times
