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M01 Q19

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13 Jul 2012, 17:44
My question is why can't you square both sides of statement 2? You would then get x> y

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14 Jul 2012, 00:52
priyankur_saha@ml.com wrote:
Is $$X \gt Y$$ ?

1. $$\sqrt{X} \gt \sqrt{Y}$$
2. $$X^2 \gt Y^2$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I do not agree OA and OE. Please provide explanation

OE is
From S1, since X and Y are under a radical, they are nonnegative. So we may safely square them and get . So, S1 is sufficient.
According to S2 and can be negative, so we may not insist that

My Question:
Why it is assumed that rt(X) is non-negative?
If x=4, rt(x) could be +2 /-2.

And based on that answer should be E. Let me know if I am doing any mistake.

(1): $$X$$ and $$Y$$ must be both positive. Also, $$\sqrt{X}$$ and $$\sqrt{Y}$$ are both positive. Squaring the given inequality, we get $$X>Y$$. Sufficient.
Note: If we don't know for sure that both sides of an inequality are positive, we are not allowed to square it. See, for example $$1 > -2$$, but $$1 > 4$$ is false.
Also, the square root of a positive number is positive (by definition)!

(2) The given inequality can be rewritten as $$X^2-Y^2>0$$ or $$(X+Y)(X-Y)>0$$. The last one states that $$X+Y$$ and $$X-Y$$ are either both positive or both negative. Therefore, both scenarios $$X > Y$$ and $$X < Y$$ are possible. Not sufficient.

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20 Jul 2012, 13:16
dmk112 wrote:
My question is why can't you square both sides of statement 2? You would then get x> y

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I presume, you mean to say that why can't we take square root of both sides of statement 2.

So here goes, the square root can lead to two values + or -. It is best explained by plugging in 2 numbers.

For example : x^2 = 9 and y^2 = 4

taking square-root, x= +3 or -3 and y = +2 or -2.

As you see here, if x=-3 and y=2 then x is not greater than y. Thus statement (ii) is insufficient and the answer is A (only statement i is sufficient).
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20 Jul 2012, 13:57
avrgmat wrote:
dmk112 wrote:
My question is why can't you square both sides of statement 2? You would then get x> y

Posted from my mobile device

I presume, you mean to say that why can't we take square root of both sides of statement 2.

So here goes, the square root can lead to two values + or -. It is best explained by plugging in 2 numbers.

For example : x^2 = 9 and y^2 = 4

taking square-root, x= +3 or -3 and y = +2 or -2.

As you see here, if x=-3 and y=2 then x is not greater than y. Thus statement (ii) is insufficient and the answer is A (only statement i is sufficient).

You can take the square root of both sides of statement (2) if you do it properly.
$$\sqrt{X^2}=|X|$$ and $$\sqrt{Y^2}=|Y|$$, so you get $$|X|>|Y|$$ and continue from here...

By definition, square root is always non-negative. But the quadratic equation $$x^2=4$$ has two roots, 2 and -2.
Taking the square root of both sides, you get $$|x|=2$$, therefore x can be either 2 or -2.
Don't confuse taking the square root of a non-negative number with finding the roots of a quadratic equation.

It is incorrect to say "the square root can lead to two values + or -". There is no such a thing in mathematics.
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Re: M01 Q19   [#permalink] 20 Jul 2012, 13:57

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