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x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =

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New post 10 Jun 2015, 04:29
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x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)


Kudos for a correct solution.

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x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 10 Jun 2015, 08:09
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Bunuel wrote:
x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)


Kudos for a correct solution.


Given: \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\)

i.e. \(2(\sqrt{x}-\sqrt{y})=\frac{\sqrt{x[}{square_root]^2-[square_root]y}^2/b}\)

i.e. \(2(\sqrt{x}-\sqrt{y})=\frac{(\sqrt{x[}{square_root]-[square_root]y})*(\sqrt{x}+\sqrt{y})/b}\)

i.e. \(2=\frac{(\sqrt{x[}{square_root]+[square_root]y})/b}\)

i.e. \(2b=(\sqrt{x}+\sqrt{y})\)

Answer: Option
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 11 Jun 2015, 03:37
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\(2\sqrt{x}-2\sqrt{y}=\frac{(x-y)}{b}\)

1. Factor 2 out:

\(2(\sqrt{x}-\sqrt{y})=\frac{(x-y)}{b}\)

2. Multiply both sides by b:

\(2b(\sqrt{x}-\sqrt{y})=x-y\)

3. Isolate 2b, by dividing both sides by \((\sqrt{x}-\sqrt{y})\):

\(2b=\frac{(x-y)}{(\sqrt{x}-\sqrt{y})}\)

4. Multiply right side by conjugate to eliminate square roots from denominator:

\(2b=\frac{(x-y)*}{(\sqrt{x}-\sqrt{y})} * (\sqrt{x}+\sqrt{y})/(\sqrt{x}+\sqrt{y})\)

\((\sqrt{x}-\sqrt{y})\) and \((\sqrt{x}+\sqrt{y})\) simplify to "x-y" and that cross cancels with numerator, hence leaving us with \((\sqrt{x}+\sqrt{y})\)

Answer:
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 15 Jun 2015, 01:42
Bunuel wrote:
x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

\(2b(\sqrt{x}-\sqrt{y})=x-y\)

\(2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}\)

Note that it is OK to divide by \(\sqrt{x}-\sqrt{y}\), since x > y , which implies that \(\sqrt{x}-\sqrt{y}\neq{0}\).

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

\(2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}\)

Cancel \(\sqrt{x}-\sqrt{y}\) in the numerator and denominator to get \(2b = \sqrt{x}+\sqrt{y}\).

The correct answer is B.
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 16 Jun 2015, 06:33
Bunuel wrote:
Bunuel wrote:
x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

\(2b(\sqrt{x}-\sqrt{y})=x-y\)

\(2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}\)

Note that it is OK to divide by \(\sqrt{x}-\sqrt{y}\), since x > y , which implies that \(\sqrt{x}-\sqrt{y}\neq{0}\).

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

\(2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}\)

Cancel \(\sqrt{x}-\sqrt{y}\) in the numerator and denominator to get \(2b = \sqrt{x}+\sqrt{y}\).

The correct answer is B.



Hello,

could you show me the difference of 2 squares? I cannot see it...
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 16 Jun 2015, 06:57
pacifist85 wrote:
Bunuel wrote:
Bunuel wrote:
x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

\(2b(\sqrt{x}-\sqrt{y})=x-y\)

\(2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}\)

Note that it is OK to divide by \(\sqrt{x}-\sqrt{y}\), since x > y , which implies that \(\sqrt{x}-\sqrt{y}\neq{0}\).

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

\(2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}\)

Cancel \(\sqrt{x}-\sqrt{y}\) in the numerator and denominator to get \(2b = \sqrt{x}+\sqrt{y}\).

The correct answer is B.



Hello,

could you show me the difference of 2 squares? I cannot see it...


x-y= (\(\sqrt{x}\) - \(\sqrt{y}\) )( \(\sqrt{x}\) + \(\sqrt{y}\) )
This is of the same identity as \(a^2 - b^2\)= (a+b)(a-b)
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 16 Jun 2015, 07:42
Well I am still missing sth.

We are at: 2b=(x−y) / (√x−√y)
The direrence of squares is this: (x-y)^2 = (x-y)(x+y). But in our case x - y is not raised to the second power..

Then I though you squared the nominator and the denominator, which could create the above mentioned difference of squares, but in the denominator (√x - √y )( √x + √y), and should also create this in the nominator (x-y) (x+y).

So, I still cannot see the difference of squares...
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =  [#permalink]

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New post 16 Jun 2015, 08:34
pacifist85 wrote:
Well I am still missing sth.

We are at: 2b=(x−y) / (√x−√y)
The direrence of squares is this: (x-y)^2 = (x-y)(x+y). But in our case x - y is not raised to the second power..

Then I though you squared the nominator and the denominator, which could create the above mentioned difference of squares, but in the denominator (√x - √y )( √x + √y), and should also create this in the nominator (x-y) (x+y).

So, I still cannot see the difference of squares...


\((a-b)^2=a^2-2ab+b^2\) is square of the difference.

\(a^2 - b^2=(a-b)(a+b)\) is difference of two squares.

In our case we have x - y, which is \((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y\).

Hope it's clear.
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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