√3+√2/√3-√2 is approximately between which two numbers?

Question

\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}  is approximately between which two numbers?

A. 3 and 4 
B. 4 and 5 
C. 5 and 8 
D. 9 and 11 
E. 11 and 13

archit1996 · Answer

Between 9 and 11
When you rationalize, the fraction becomes 5+2√6 
Since √6 is less than 2.5 (as 2.5^2 = 6.25), the whole expression would be somewhere around 10. So the answer would be between 9 and 11.

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Sirakri · Answer

Rationalising, we get.

5+2\sqrt{2}\sqrt{3} 
5+2(~1.4*~1.7)=5+2(~2.38)=~9.76

So, 9 & 11.

D.

EgmatQuantExpert · Answer

Hey,

First of all, we need to rationalize the given fraction.

\frac{√3+√2}{√3-√2}

So we will multiply the numerator and denominator by the conjugate of √3-√2, which is √3+√2.

Thus, we will get -

\frac{(√3+√2)(√3+√2)}{(√3-√2)(√3+√2)} 
= \frac{(√3+√2)^2}{(√3)^2-(√2)^2} 
= \frac{(√3)^2+(√2)^2+2.√3.√2}{(√3)^2-(√2)^2} 
=  \frac{(5+2.√6)}{(3-2)}

Now √6 lies between 2 and 3  
i.e.   2 <√6 <3  
Taking the extreme cases, we can conclude that the fraction will lie between -

5 + 2*2<\frac{√3+√2}{√3-√2}<5 + 2*3

9<\frac{√3+√2}{√3-√2}< 11

Thus the correct answer is  Option D

Thanks,
Saquib
Quant Expert
e-GMAT

To practise ten 700+ Level Number Properties Questions attempt the  The E-GMAT Number Properties Knockout

https://e-gmat.com/blogs/wp-content/uploads/2016/12/NP-Free-Trial-e1481107541356.png

raeann105 · Answer

MathRevolution  Shouldn't this be written as (√3+√2)/(√3-√2) or  \frac{√3+√2}{√3-√2} ? What you have written appears to be a completely different math problem or am I missing something?

Sirakri · Answer

You are right mate!

MathRevolution · Answer

==>From  √3+√2/√3-√2=(√3+√2)^2/(√3-√2)( √3+√2)=3+2+2√6/3-2=5+2√6=5+2(2.xxx)=9.xxx ,

the answer is D.
Answer: D

Abhishek009 · Answer

\frac{(√3+√2)}{(√3-√2)}

=  \frac{(√3+√2)(√3+√2)}{(√3-√2)(√3+√2)}

=  \frac{(√3^2 + √2^2 + 2√3√2)}{( 3 - 2 )}

=  3 + 2 + 2√6

=  5 + 2√6

=  5 + 2(2.45)

=  5 + 4.90

=  9.90

Hence, the correct answer must be (D) 9 and 11

BrushMyQuant · Answer

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=WS0rQP1mzok

We need to find the approximate value of  \frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}

\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}

Now to simplify this we will multiply the numerator and the denominator with the conjugate of the denominator, which is  \sqrt3 + \sqrt2

=>  \frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}  =  \frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}  *  \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2}  =  \frac{(\sqrt3+\sqrt2) * (\sqrt3 + \sqrt2)}{(\sqrt3-\sqrt2) * (\sqrt3 + \sqrt2)}

Now, numerator becomes  (a+b)^2 = a^2 + 2ab + b^2  and the denominator becomes (a-b) * (a+b) =  a^2 - b^2

=>  \frac{(\sqrt3+\sqrt2) * (\sqrt3 + \sqrt2)}{(\sqrt3-\sqrt2) * (\sqrt3 + \sqrt2)}  =  \frac{(\sqrt3)^2 + 2 *\sqrt3*\sqrt2 + (\sqrt2)^2 }{(\sqrt3)^2 - (\sqrt2)^2}  =  \frac{3 + 2\sqrt6 + 2 }{ 1}  = 5 +  2\sqrt6  ~ 5 + 2*2.45 = 5 + 4.9 = 9.9
=> Between 9 and 11

So,  Answer will be D 
Hope it helps!

Watch the following video to learn How to Rationalize Roots

https://www.youtube.com/watch?v=LhR_gXt8CLw

√3+√2/√3-√2 is approximately between which two numbers?

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√3+√2/√3-√2 is approximately between which two numbers?

Prep Toolkit

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