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1) sqrt(x)>y from this it is clear that x must be +ve.. x=4 y=1 x>y x=1/4 y=1/3 x<y this is where i disagree with you. if x=1/4 then sq.root(x) can be 1/2 or -1/2. For satisfying the condition sq.root(x)>y, y < -1/2. That would automatically make x>y. That is why i think A should be the answer.

insuffcient

2) x^3>y x *x^2 >y (x^2 is always +ve)

x=4 y=1 x>y x=-1/2 y=-1/4 x<y insuffcieint

combine. x must be +ve and not <1 and >0 e.g x=1/4 y=1/3 (Statement 2 fails)

so x>y (any x>1 such that sqrt(x)>y)

C

Any other short cuts..

In GMAT universe, square roots are always positive.

Not quite. I dont need fraction values here. we know x > 0.

Taking your example, if x=1/4 then sqrt(x) = 1/2 or -1/2. To satisfy the statement 1 sqrt(x) > y, you need to account for all possible values of sqrt(x) and for that y has to be smaller than -1/2. So you have +ve x and -ve y. Thus sufficient.

Not quite. I dont need fraction values here. we know x > 0.

Taking your example, if x=1/4 then sqrt(x) = 1/2 or -1/2. To satisfy the statement 1 sqrt(x) > y, you need to account for all possible values of sqrt(x) and for that y has to be smaller than -1/2. So you have +ve x and -ve y. Thus sufficient.

Bushan,

Let me put this way. Here given equation is sqrt(x)>y I pick value for sqrt(x)=1/2 and y=1/3 such that it satisfies sqrt(x)>y

Now I will find the value for x = sqrt(x)*sqrt(x) =1/4 y=1/3 x<y
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Here given equation is sqrt(x)>y I pick value for sqrt(x)=1/2 and y=1/3 such that it satisfies sqrt(x)>y If you pick sqrt(x)=1/2, then by default you have to consider -1/2. It is the charecteristic property of the square root function that it will have two values a +ve and a -ve. You can't consider the +ve alone unless it is categoricaly mentioned. So if you have a situation where sqrt(x) > y, then y has to be less than the -ve root. But here i guess the answer will be C. Zoink and Set009 are so sure about the GMAT rule for roots. Now I will find the value for x = sqrt(x)*sqrt(x) =1/4 y=1/3 x<y

Zoink and Sset009.. thanx for bringing that up.

Last edited by bhushangiri on 28 Aug 2008, 13:35, edited 1 time in total.

(1) If sqrt(x) > y, x > y if x is greater than or equal to 1, since for such values of x, x is greater than or equal to sqrt(x) . If 0< x < 1, sqrt(x) > x, so the relationship between x and y cannot be determined. (2) If x^3 > y, x > y if 0 < x < 1, since for such values of x , x > x^3. However, if x > 1, x < x^3, so the relationship between x and y is unclear.

(T) For x not be to greater than y, (1) tells us that 0<x<1, but (2) tells us that x is greater than 1, a contradiction. Thus x must be greater than y.

In GMAT universe, square roots are always positive.

No. In GMAT universe, squares are always positive. Square roots can be -ve.

NEVER, square roots are +ve by definition. pg 114 OG 11

The OG says the exact opposite of that. If you read that page, it says "Every positive number n has two square roots, one positive and the other negative." It is only when dealing with the square root symbol (radicals) that you can never have a negative result:

\(\sqrt{x} \geq 0\) for all x, because, by definition \(\sqrt{x}\) is equal to the non-negative square root of x only.

Note that radicals are not always positive; they can be zero.
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