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x - y > 0 and x > 1 and √(-)=√x

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Director
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x - y > 0 and x > 1 and √(-)=√x [#permalink] New post 16 Aug 2018, 09:03
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Question Stats:

60% (01:24) correct 40% (01:32) wrong based on 15 sessions
\(x - y > 0\) and \(x > 1\) and \(\sqrt{(x-y)}=\sqrt{x}-1\)

Quantity A
Quantity B
\(Y\)
\(2 \sqrt{x}-1\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA
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Re: x - y > 0 and x > 1 and √(-)=√−1 [#permalink] New post 17 Aug 2018, 06:48
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Expert's post
amorphous wrote:
\(x - y > 0\) and \(x > 1\) and \(\sqrt{(x-y)}=\sqrt{-1}\)

Quantity A
Quantity B
\(Y\)
\(2 \sqrt{-1}\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


I think this question is missing something.
As it stands, we're told that x - y > 0. In other words, x - y is POSITIVE

Then we're told that \(\sqrt{(x-y)}=\sqrt{-1}\)
This means x - y = -1
However, we earlier learned that x - y is POSITIVE

Cheers,
Brent
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Director
Director
User avatar
Joined: 07 Jan 2018
Posts: 553
Followers: 4

Kudos [?]: 477 [0], given: 84

CAT Tests
Re: x - y > 0 and x > 1 and √(-)=√−1 [#permalink] New post 02 Sep 2018, 20:49
GreenlightTestPrep wrote:
amorphous wrote:
\(x - y > 0\) and \(x > 1\) and \(\sqrt{(x-y)}=\sqrt{-1}\)

Quantity A
Quantity B
\(Y\)
\(2 \sqrt{-1}\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


I think this question is missing something.
As it stands, we're told that x - y > 0. In other words, x - y is POSITIVE

Then we're told that \(\sqrt{(x-y)}=\sqrt{-1}\)
This means x - y = -1
However, we earlier learned that x - y is POSITIVE

Cheers,
Brent


yes the question was incomplete, I have changed it.
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Posts: 199
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Kudos [?]: 63 [0], given: 1

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Re: x - y > 0 and x > 1 and √(-)=√x [#permalink] New post 21 Oct 2018, 01:30
Expert's post
amorphous wrote:
\(x - y > 0\) and \(x > 1\) and \(\sqrt{(x-y)}=\sqrt{x}-1\)

Quantity A
Quantity B
\(Y\)
\(2 \sqrt{x}-1\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


x-y>0 ; X>1 ; \(\sqrt{(x-y)}=\sqrt{x}-1\)

Let us square \(\sqrt{(x-y)}=\sqrt{x}-1\)..
\(\sqrt{(x-y)}^2=(\sqrt{x}-1)^2........ x-y=x+1-2√x....y=2√x-1.....A=B\)

C
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: x - y > 0 and x > 1 and √(-)=√x   [#permalink] 21 Oct 2018, 01:30
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