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GMATClub Test M12 Q1 - Faster method?

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Intern
Joined: 29 Nov 2009
Posts: 20

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

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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!

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

Kudos [?]: 19 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133020 [0], given: 12402

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

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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: $$\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$$

_________________

Kudos [?]: 133020 [0], given: 12402

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

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12 Feb 2010, 17:08

Is it gmatclub test with the wrong answer? hmmmm

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Intern
Joined: 29 Nov 2009
Posts: 20

Kudos [?]: 19 [0], given: 5

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

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12 Feb 2010, 17:21
chix475ntu wrote:

Is it gmatclub test with the wrong answer? hmmmm

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

Kudos [?]: 19 [0], given: 5

Intern
Joined: 29 Nov 2009
Posts: 20

Kudos [?]: 19 [0], given: 5

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

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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: $$\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$$

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

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Senior Manager
Joined: 01 Feb 2010
Posts: 251

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

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13 Feb 2010, 10:52

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

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

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

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

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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!

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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