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GMATClub Test M12 Q1 - Faster method?

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GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 12 Feb 2010, 16:37
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

[Reveal] Spoiler:
B
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)


Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!

edit: corrected answer choice. whoops.

Last edited by brentbrent on 12 Feb 2010, 17:22, edited 1 time in total.

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 12 Feb 2010, 17:00
brentbrent wrote:
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

[Reveal] Spoiler:
D
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)


Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!


Answer can not be D, it should be B. The question is about the application of formula: \(a^2-b^2=(a-b)(a+b)\). Basically what we want to do is to make denominator 1, as no answer choice is in the form of fraction.

How can we do that?

Multiply \(\frac{1}{2+\sqrt{3}}\) by \(\frac{2-\sqrt{3}}{2-\sqrt{3}}\), which is 1, so that won't affect the value of our fraction. We'll get: \(\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)

Answer: B.
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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 12 Feb 2010, 17:08
Answer should be B.

Is it gmatclub test with the wrong answer? hmmmm :shock:

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 12 Feb 2010, 17:21
chix475ntu wrote:
Answer should be B.

Is it gmatclub test with the wrong answer? hmmmm :shock:



Whoops.
Answer is B. I have to edit that, gmat club isn't wrong =-)

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 12 Feb 2010, 17:25
Bunuel wrote:
brentbrent wrote:
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

[Reveal] Spoiler:
D
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)


Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!


Answer can not be D, it should be B. The question is about the application of formula: \(a^2-b^2=(a-b)(a+b)\). Basically what we want to do is to make denominator 1, as no answer choice is in the form of fraction.

How can we do that?

Multiply \(\frac{1}{2+\sqrt{3}}\) by \(\frac{2-\sqrt{3}}{2-\sqrt{3}}\), which is 1, so that won't affect the value of our fraction. We'll get: \(\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)

Answer: B.

Thanks Bunnel. I didn't even think about that formula..sheesh.
cheers

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 13 Feb 2010, 10:52
Answer is B

Multiply numerator & denominator by 2-\sqrt{3}.

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 14 Feb 2010, 01:36
20 sec question and answer must be B.

multiply numerator and denominator by 2-\sqrt{3}

Cheers!
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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 16 Feb 2010, 14:05
How do you go from 2-[square_root]3/4-3 to the final answer?

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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 17 Feb 2010, 01:30
Crushit wrote:
How do you go from 2-[square_root]3/4-3 to the final answer?


just simplify it

2 - sqrt(3) / 4 - 3

= 2 - sqrt(3)/1

= 2 - sqrt(3) which is the answer
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Re: GMATClub Test M12 Q1 - Faster method? [#permalink]

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New post 06 Mar 2010, 11:04
brentbrent wrote:
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

[Reveal] Spoiler:
B
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)


Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!

edit: corrected answer choice. whoops.




Try this method instead....

we knw sqaure root of 3 is 1.732

So the equation becomes 1/(2+1.732)
that gives 1/(3.732) or may be rounded off to 1/4 or 0.25

Now working with options only B gives 2-1.732 that would give answer as 0.25.


Hope this helps.
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Re: GMATClub Test M12 Q1 - Faster method?   [#permalink] 06 Mar 2010, 11:04
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