If a = 0.1039, then the value of √4a² - 4a + 1 + 3a is:

3√5 + √125 = 17.88

⟹ 3√5 + √25 x 5 = 17.88

⟹ 3√5 + 5√5 = 17.88

⟹ 8√5 = 17.88

⟹ √5 = 2.235

∴ √80 + 6√5 = √16 x 5 + 6√5

   = 4√5 + 6√5

   = 10√5 = (10 x 2.235) = 22.35

Let	x	=	√162
	√128		x

Then x² = √128 x 162

   = √64 x 2 x 18 x 9

   = √8² x 6² x 3²

   = 8 x 6 x 3

   = 144.

∴ x = √144 = 12.

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

❨	√3 -	1	❩	2	= (√3)2 +	❨	1	❩	2	- 2 x √3 x	1
		√3					√3				√3

= 3 +	1	- 2
	3

= 1 +	1
	3

=	4
	3

(7 + 3√5)(7 - 3√5)

=

(7)2 - (3√5)2

= √49 - 45  = √4  = 2.

   1|1.5625( 1.25
    |1
    |-------
  22| 56
    | 44
    |-------
 245| 1225
    | 1225
    |-------
    |    X
    |-------     

∴ √1.5625 = 1.25.

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 =	❨	5 x	3	❩	:	❨	8 x	3	❩	= 3 :	24
			5					5			5

⟹ A : B : C = 2 : 3 :	24	= 10 : 15 : 24
	5

⟹ B =	❨	98 x	15	❩	= 30.
			49

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

⟹	❨	120	x 7x	❩	and	❨	110	x 8x	❩
		100					100

⟹	42x	and	44x
	5		5

∴ The required ratio =	❨	42x	:	44x	❩	= 21 : 22.
		5		5

Let 40% of A =	2	B
	3

Then,	40A	=	2B
	100		3

⟹	2A	=	2B
	5		3

⟹	A	=	❨	2	x	5	❩	=	5
	B			3		2			3

∴ A : B = 5 : 3.

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then,	2x + 4000	=	40
	3x + 4000		57

⟹ 57(2x + 4000) = 40(3x + 4000)

⟹ 6x = 68,000

⟹ 3x = 34,000

Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.