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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))

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Director
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Updated on: 10 Jun 2016, 15:00
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Difficulty:

25% (medium)

Question Stats:

73% (01:03) correct 27% (01:54) wrong based on 244 sessions

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$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

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Md. Abdur Rakib

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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

Originally posted by AbdurRakib on 10 Jun 2016, 13:05.
Last edited by Bunuel on 10 Jun 2016, 15:00, edited 1 time in total.
Renamed the topic and edited the question.
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Updated on: 09 Jan 2018, 07:13
1
The trick behind this question is to use the answer choices to guide your mathematics. (In other words, don't do math just because you can... do math because it gets you closer to your target!)

You should notice immediately that the answer choices are much simpler. Three of the five answer choices contain no fractions at all. None of the answer choices contain $$\sqrt{6}$$. Use this to your advantage to help you think about the question. One quick way to get rid of both the \sqrt{6}[/m] and the "fractions inside of fractions" issue is to do what I call "Multiply by 1". In this case, multiply the original expression by $$\sqrt{6}/\sqrt{6}$$. Since $$\sqrt{6}/\sqrt{6}=1$$, we don't actually change the value; we just change the shape of the expression. Thus,

$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}*(\frac{\sqrt{6}}{\sqrt{6}})=\frac{6}{\sqrt{3}+\sqrt{2}}$$

Now the expression that remains is starting to look more like the answer choices, but we still need to get rid of the compound denominator $$(\sqrt{3}+\sqrt{2})$$. Once again, look to the answer choices for clues. Both answer choices A and E contain a $$6$$. But it should be clear that answer choice A cannot be an option. $$(\sqrt{3}+\sqrt{2})$$ does not equal 5. So, let's try to turn what we have into answer choice E.

Answer choice E contains a factor of $$(\sqrt{3}-\sqrt{2})$$, which should immediately get us thinking about the possibility of difference of squares. Multiplying our reduced expression by "1" again gives us:

$$(\frac{6}{\sqrt{3}+\sqrt{2}})*(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})$$

Using the concept of difference of squares, we can quickly see that the denominator of this fraction simplifies down completely. Watch this "disappearing math" trick:

$$(\sqrt{3}+\sqrt{2})*(\sqrt{3}-\sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3-2 = 1$$

What began as a "Mathugly" arithmetic issue quickly simplifies down to $$6\sqrt{3}-6\sqrt{2}$$.

The answer is E.

Addendum: As with many quantitative questions on the GMAT, there is "more than one way to skin a CAT" for this question (pun intended.) We can also use the answer choices not just as targets that guide our math, but as leverage for approximating. Watch this. If we recognize that $$\sqrt{3}\approx{1.7}$$ and $$\sqrt{2}\approx{1.4}$$, then the we can approximate our answer very quickly:
$$\frac{6}{\sqrt{3}+\sqrt{2}}\approx{\frac{6}{1.4+1.7}}\approx{\frac{6}{3.1}}\approx{2}$$

Now, we look at the answer choices to see which one is roughly equal to 2.

(A) $$\frac{6}{5}$$ <-- quickly eliminated. Too small.

(B) $$\frac{5}{6}$$ <-- quickly eliminated. Too small.

(C) $$5$$ <-- quickly eliminated. Too big.

(D) $$2\sqrt{3} + 3\sqrt{2}$$ <-- quickly eliminated. Too big.

(E) $$6\sqrt{3} - 6\sqrt{2}$$ <-- The right answer. $$6(\sqrt{3} - \sqrt{2})\approx{6(1.7-1.4)}\approx{2}$$

Anyway you look at it, the answer is still E.
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Aaron J. Pond
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Originally posted by AaronPond on 08 Jan 2018, 17:04.
Last edited by AaronPond on 09 Jan 2018, 07:13, edited 1 time in total.
##### General Discussion
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10 Jun 2016, 15:13
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}})}=\frac{\sqrt{6}*\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{6}{\sqrt{3}+\sqrt{2}}$$.

Multiply by $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$:

$$\frac{6}{\sqrt{3}+\sqrt{2}}*\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{6*(\sqrt{3}-\sqrt{2})}{3-2}=6*\sqrt{3}-6\sqrt{2}$$.

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25 Jul 2017, 08:45
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

Nice question!

Here's another approach:

Let's first deal with the denominator: 1/√2 + 1/√3 = √3/√6 + √2/√6
= (√3 + √2)/√6

So....$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$ √6/[(√3 + √2)/√6]

= (√6)(√6)/(√3 + √2)

= 6/(√3 + √2)

Now we can use the fact that √3 ≈ 1.7 and √2 ≈ 1.4 to get....
≈ 6/(1.7 + 1.4)
≈ 6/(3.1)

Since 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2

Now check the answer choices to see which one evaluates to be a number a bit smaller than 2

A. $$\frac{6}{5}$$ ELIMINATE

B. $$\frac{5}{6}$$ ELIMINATE

C. 5 ELIMINATE

D. $$2\sqrt{3} +3\sqrt{2}$$ ≈ 2(1.7) + 3(1.4) ≈ a number that's bigger than 6 ELIMINATE

E. $$6\sqrt{3}-6\sqrt{2}$$[/quote] = 6(√3 - √2) ≈ 6(1.7 - 1.4) ≈ 6(0.3) ≈ 1.8 PERFECT!

Cheers,
Brent
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08 Jan 2018, 16:36
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$

(A) $$\frac{6}{5}$$

(B) $$\frac{5}{6}$$

(C) $$5$$

(D) $$2\sqrt{3} + 3\sqrt{2}$$

(E) $$6\sqrt{3} - 6\sqrt{2}$$
_________________

Aaron J. Pond
Veritas Prep Elite-Level Instructor

Hit "+1 Kudos" if my post helped you understand the GMAT better.
Look me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.

Math Expert
Joined: 02 Sep 2009
Posts: 49298

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08 Jan 2018, 20:33
AaronPond wrote:
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$

(A) $$\frac{6}{5}$$

(B) $$\frac{5}{6}$$

(C) $$5$$

(D) $$2\sqrt{3} + 3\sqrt{2}$$

(E) $$6\sqrt{3} - 6\sqrt{2}$$

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Merging topics.
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Re: 6^(1/2)/(1/2^(1/2)+1/3^(1/2)) &nbs [#permalink] 08 Jan 2018, 20:33
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