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# M01 Q19

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M01 Q19 [#permalink]

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24 Feb 2009, 20:38
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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.
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Re: m - 01 Q - 19 [#permalink]

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24 Feb 2009, 20:43
priyankur_saha@ml.com wrote:
I do not agree OA and OE. Please provide explanation
Attachment:
math.JPG

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.

GMAT considers only +ve value for Sqrt()

Sqrt(x) --> lead only one +ve solution.

X^2=4 then both +ve and -ve are valid solutions.
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Re: M01 Q19 [#permalink]

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28 Apr 2010, 20:30
Please elaborate on that. You mean GMAT considers all radical terms positive?

For example, $$\sqrt{4}$$ can only be 2, whereas $$x^2=4$$ x can be +/- 2?

Then why do we have to keep remembering that the sq rt of something can either be positive or negative?
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Re: M01 Q19 [#permalink]

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28 Apr 2010, 21:00
I marked C first but later I realised that OA should be A.

S1: Suff
a. If x=1/4 and y=1/9 >>>> x>y
b. If x=9 and y=4 >>>> x>y

S2: X and Y can be +ve or -ve. Insuff.
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Re: M01 Q19 [#permalink]

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07 Jul 2010, 03:44
Quote:
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.

Reply:

If you graph the square root of a number then the graph will always be on the positive side. This is a basic principle of math that we learned from pre-calculus course or even earlier. Even Calculus text-book should have them. This is how i see why square root is always positive.

Another way to see this, rt ( -4) doesn't exist because you can not square a number and then get a negative number. I believe that GMAT doesn't want you to break it down from rt(4) to +2/-2. They want you to take it as is. If its rt(4) then you should take it as x=4
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Re: M01 Q19 [#permalink]

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07 Jul 2010, 04:34
You can see the OA by clicking "Reveal" in the spoiler under the question.
RenukaD wrote:
What is the OA?

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Re: M01 Q19 [#permalink]

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07 Jul 2010, 04:40
dzyubam wrote:
You can see the OA by clicking "Reveal" in the spoiler under the question.
RenukaD wrote:
What is the OA?

Thanks dzyubam . my answer was also A but ,looking at posts above, which says "GMAT does not consider -ve values of root" so by this are we trying to say in this case "C" is sufficient condition ? If not then in what scenario we should consider only positive values of roots.

Thanks in advance
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Re: M01 Q19 [#permalink]

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07 Jul 2010, 04:45
No, the answer remains A for this question. Here's a quote that should help you understand when only positive roots are considered:
Quote:
Sqrt(x) --> lead only one +ve solution.

X^2=4 then both +ve and -ve are valid solutions.

GMAT deals with only positive numbers under the square root sign.
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Re: M01 Q19 [#permalink]

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07 Jul 2010, 04:51
dzyubam wrote:
No, the answer remains A for this question. Here's a quote that should help you understand when only positive roots are considered:
Quote:
Sqrt(x) --> lead only one +ve solution.

X^2=4 then both +ve and -ve are valid solutions.

GMAT deals with only positive numbers under the square root sign.

Thanks again dzyubam Now its clear to me....
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Re: M01 Q19 [#permalink]

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07 Jul 2010, 13:32
I agree - A is correct in my opinion.
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Re: M01 Q19 [#permalink]

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08 Jul 2010, 00:08
Is X>Y?

1) X^(1/2) > Y^(1/2)
- Because X and Y are under a square-root they can only be positive, or imaginery numbers. thus we can conclude that statment 1 is sufficient.

2) X^2 > Y^2
- This statement tells us that the absolute value of X is greater than Y, but we cannot determine that X is greater than Y, thus this statement is insufficient. For instance:
X=3, -3
Y=2, -2
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Re: M01 Q19 [#permalink]

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08 Jul 2010, 16:41
another way to solve this is to solve the inequality -

for the first one: square both sides and you get X> Y .... simple and sufficient.

for the second one: X or Y can be +/- hence not sufficient.

A...
thanks
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Re: M01 Q19 [#permalink]

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08 Jul 2010, 17:11
Answer is A

sqrt(x) > sqrt(y)

If X=16, Y=4 then sqrt(x) > sqrt (y)

If X=1/4, y=1/16 then sqrt(x) > sqrt(y)

GMAT considers +ve sign for root of numbers. So, this condition is sufficient. Correct answer is A

Condition 2

X2 > Y2

for X<Y, X=-4 and Y=-2, X2>Y2
for X>Y, X=4 and Y=2, X2>Y2

So, Condition 2 is not sufficient
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Re: M01 Q19 [#permalink]

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11 Jul 2011, 06:55
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A is sufficient. You square both sides and you get X > Y.

But B otoh is not sufficient. Eg: 25 > 16, but -5 < 4.
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Re: M01 Q19 [#permalink]

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11 Jul 2011, 10:18
A for me too..
Negative Sq. rt. leads to an imaginary value which we used to refer as 'iota' in school..
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Re: M01 Q19 [#permalink]

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14 Jul 2011, 07:14
(I) if \sqrt{x}>\sqrt{y}
x>y...\sqrt{x} can only be $$>=$$ 0
x,y are > 0....hence x > y
Sufficient

(II) if x^2 > y^2
x^2 - y^2 >0
(x+y) or (x-y) > 0
both brackets need to be positive or negative.

Both brackets positive:
x=4, y= -2;
(4-2)(4--2)>0
from here, we have (4-2)(4+2)>0
And x(4) > y.....response is YES
sufficient

Both brackets Negative:
x=-4; y= -2
(-4 -2) and (-4 +2)
from here, we have: (-4-2)(-4+2)>0
And x(-4) < y(-2).....response is NO
insufficient
So, II is INSUFFICIENT

OA is A.
SImply put, test the pair of numbers with different signs:
(i) x(4) and y(2)
(ii) x(-4) and y(-2) ...
and you will realize the insufficiency in II.
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Re: M01 Q19 [#permalink]

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06 Sep 2011, 07:00
if sqrtX=sqrtY=0 then not sufficient. Am I wron?
suppose x=0 and y=0.
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Re: M01 Q19 [#permalink]

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07 Sep 2011, 13:48
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Re: M01 Q19 [#permalink]

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13 Jul 2012, 04:12
thanatoz wrote:
Please elaborate on that. You mean GMAT considers all radical terms positive?

For example, $$\sqrt{4}$$ can only be 2, whereas $$x^2=4$$ x can be +/- 2?

Then why do we have to keep remembering that the sq rt of something can either be positive or negative?

I know this might come as a source of confusion so let me clarify it well:

The square root symbol means, in fact, the +ve square root of. That's a definition and sometimes we just drop the word "positive" to say simply "the square root of".

Regarding your equation $$x^2=4$$ x can be +/- 2, we have 2 answers because of the x^2 and nothing else so that gives 2 answers: one is the positive square root of 4 (which is 2) and the other is the negative of the square root of 4 (which is -2).

Hope this helps...
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Re: M01 Q19 [#permalink]

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13 Jul 2012, 05:02
A

x = -3 and y = 2

9> 4
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Re: M01 Q19   [#permalink] 13 Jul 2012, 05:02

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