The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:

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

∴ √1.5625 = 1.25.

√4a² - 4a + 1 + 3a = √(1)² + (2a)² - 2 x 1 x 2a + 3a

   = √(1 - 2a)² + 3a

   = (1 - 2a) + 3a

   = (1 + a)

   = (1 + 0.1039)

   = 1.1039

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

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.

Given ratio = ½ : ⅔ : ¾ = 6 : 8 : 9.

∴ 1st part = Rs.	❨	782 x	6	❩	= Rs. 204
			23

	4	A	=	2	B
	15			5

⟹ A =	❨	2	x	15	❩B
		5		4

⟹ A =	3	B
	2

⟹	A	=	3
	B		2

⟹ A : B = 3 : 2.

∴ B's share = Rs.	❨	1210 x	2	❩	= Rs. 484.
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