What is the value of √99+√33/√99-√33 + √99-√33/√99+√33 ?

Question

What is the value of  \frac{√99+√33}{√99-√33} + \frac{√99-√33}{√99+√33}  ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

TestPrepUnlimited · Answer

Combine the fractions:

The denominator is  (\sqrt{99} - \sqrt{33})(\sqrt{99} + \sqrt{33}) = 99 - 33 = 66

The numerator is  (\sqrt{99} + \sqrt{33})^2 + (\sqrt{99} - \sqrt{33})^2 = 99 + 33 + 2*\sqrt{99*33} + 99 + 33 - 2*\sqrt{99*33} = 99 + 33 + 99 + 33

Finally we have  \frac{33 * 8 }{ 66} = 4

Ans: E

CrackverbalGMAT · Answer

Let  \sqrt{99}  = a

Let  \sqrt{33}  = b

The question therefore is  \frac{a + b}{a - b}  +  \frac{a - b}{a + b}  =  \frac{(a + b)^2 + (a - b)^2}{a^2 - b^2}

=  \frac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2 }{a^2 - b^2}  =  \frac{2a^2 + 2b^2 }{a^2 - b^2}  =  \frac{2(a^2 + b^2) }{a^2 - b^2}

=  \frac{2(\sqrt{99}^2 + \sqrt{33}^2) }{\sqrt{99}^2 - \sqrt{33}^2}

Option E

Arun Kumar

=  \frac{2(99 + 33)}{99 - 33}

=  \frac{2 * 132}{66}

= 4

nick1816 · Answer

√99 ~ 10
√33 ~ 5.8 ( √25 < √33 < √36)

\frac{√99+√33}{√99-√33} = \frac{10+5.8}{10-5.8} = \frac{15.8}{4.2} > 3

E

IanStewart · Answer

I did this the way nick did above (since √99 ~ 10 and √33 ~ 6, the sum is roughly 16/4 + 4/16, and only E makes any sense). I normally wouldn't do that for this type of question, but the first few answer choices are obviously wrong -- we're adding two positive numbers, and the first fraction is clearly larger than 1, so A and B are impossible, and since the first fraction at a glance appears to be considerably larger than 1, an estimate seemed like it might produce an answer very quickly.

The standard way to do this kind of question is by 'rationalizing denominators' -- using the difference of squares pattern to get the square roots out of each denominator. To do that, we could multiply the first fraction on top and bottom by √99 + √33, and the second on top and bottom by √99 - √33. But it is much easier to factor out √33 first, so we don't need to work with large numbers; e.g. with the first fraction:

(√99+√33)/(√99-√33) = √33(√3 + 1)/√33 (√3 - 1) = (√3 + 1)/(√3 - 1)

and to rationalize this, we multiply by √3 + 1 on top and bottom, and get (√3 + 1)^2/2. Similarly the second fraction is equal to (√3 - 1)^2/2. So their sum is

(√3 + 1)^2/2 + (√3 - 1)^2/2 = 1/2(3 + 1 + 2√3) + 1/2(3 + 1 - 2√3) = 1/2(4 + 4) = 4

You could also avoid rationalizing altogether if you noticed we're adding reciprocals, so we're doing this:

x/y + y/x = (x^2 + y^2)/xy

where x = √99 + √33, and y = √99 - √33. Then just plugging in gives the answer, though again factoring makes the calculation much simpler.

raunakd11 · Answer

Hello guys!

This took me  less than 20 seconds  to solve it. So, do look for it!

\frac{√99+√33}{√99-√33} + \frac{√99-√33}{√99+√33}

\sqrt{100}  is quite near to  \sqrt{99} , and  \sqrt{36}  is quite near to  \sqrt{33} ,

\sqrt{100} = 10 and  \sqrt{36} = 6, this value can come down by 0.3 appx.

\frac{10+6}{10-6}  +  \frac{10-6}{10+6} 
 \frac{16}{4}  +  \frac{4}{16} 
4 + 1/4,

We can stop here, we don't need to go further on,

As the initial part,  \frac{10+6}{10-6} = appx. 4 or less than it, which is greater than all the options. So, eliminate all the options except E.

Official Answer:-  Option E

Thank you!

Regards,
Raunak Damle

BongBideshini · Answer

\sqrt{99}=\sqrt{33}\sqrt{3} 
 \frac{\sqrt{99}+\sqrt{33}}{\sqrt{99}-\sqrt{33}} + \frac{\sqrt{99}-\sqrt{33}}{\sqrt{99}+\sqrt{33}}  
 = \frac{\sqrt{3}+1}{\sqrt{3}-1} + \frac{\sqrt{3}-1}{\sqrt{3}+1}  (Taking  √33  common)
 =\frac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}

=\frac{(4+2\sqrt{3})+(4-2\sqrt{3})}{3-1}

= \frac{8}{2}=4 
 Answer: E

What is the value of √99+√33/√99-√33 + √99-√33/√99+√33 ?

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What is the value of √99+√33/√99-√33 + √99-√33/√99+√33 ?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards