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x - y > 0 and x > 1 and √(-)=√x

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amorphous
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x - y > 0 and x > 1 and √(-)=√x [#permalink] New post   16 Aug 2018, 09:03
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\(x - y > 0\) 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.
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C
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GreenlightTestPrep
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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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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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amorphous
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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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chetan2u
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Re: x - y > 0 and x > 1 and √(-)=√x [#permalink] New post   21 Oct 2018, 01:30
Expert Reply
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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gmatclubot
Re: x - y > 0 and x > 1 and √(-)=√x [#permalink]
21 Oct 2018, 01:30
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