8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =	❨	7x -	7	x 9	❩	litres =	❨	7x -	21	❩ litres.
			12						4

Quantity of B in mixture left =	❨	5x -	5	x 9	❩	litres =	❨	5x -	15	❩ litres.
			12						4

∴	❨ 7x - 21 ❩ 4	=	7
	❨ 5x - 15 ❩ + 9 4		9

⟹	28x - 21	=	7
	20x + 21		9

⟹ 252x - 189 = 140x + 147

⟹ 112x = 336

⟹ x = 3.

So, the can contained 21 litres of A.

Let C.P. of 1 litre milk be Re. 1.

S.P. of 1 litre of mixture = Re.1, Gain =	50	%.
	3

∴ C.P. of 1 litre of mixture =	❨	100 x	3	x 1	❩	=	6
			350				7

By the rule of alligation, we have:

C.P. of 1 litre of water C.P. of 1 litre of milk
0	Mean Price Re. 6 7	Re. 1
1 7		6 7

∴ Ratio of water and milk =	1	:	6	= 1 : 6.
	7		7

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture =	❨	3 -	3x	+ x	❩	litres
			8

Quantity of syrup in new mixture =	❨	5 -	5x	❩	litres
			8

∴	❨	3 -	3x	+ x	❩	=	❨	5 -	5x	❩
			8						8

⟹ 5x + 24 = 40 - 5x

⟹ 10x = 16

⟹ x =	8	.
	5

So, part of the mixture replaced =	❨	8	x	1	❩	=	1	.
		5		8			5

By the rule of alligation, we have:

Strength of first jar Strength of 2nd jar
40%	Mean Strength 26%	19%
7		14

So, ratio of 1^st and 2^nd quantities = 7 : 14 = 1 : 2

∴ Required quantity replaced =	2
	3

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.

C.P. of 1 kg of the mixture = Rs.	❨	100	x 68.20	❩	= Rs. 62.
		110

By the rule of alligation, we have:

Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind.
Rs. 60	Mean Price Rs. 62	Rs. 65
3		2

∴ Required ratio = 3 : 2.

B runs	45	m in 6 sec.
	2

∴ B covers 300 m in	❨	6 x	2	x 300	❩sec	= 80 sec.
			45

B runs 35 m in 7 sec.

∴ B covers 200 m in	❨	7	x 200	❩	= 40 sec.
		35

B's time over the course = 40 sec.

∴ A's time over the course (40 - 7) sec = 33 sec.

A : B = 100 : 90.

A : C = 100 : 72.

B : C =	B	x	A	=	90	x	100	=	90	.
	A		C		100		72		72

When B runs 90 m, C runs 72 m.

When B runs 100 m, C runs	❨	72	x 100	❩m	= 80 m.
		90

∴ B can give C 20 m.

When B runs 25 m, A runs	45	m.
	2

When B runs 1000 m, A runs	❨	45	x	1	x 1000	❩m	= 900 m.
		2		25

∴ B beats A by 100 m.

A : B = 200 : 169.

A : C = 200 : 182.

C

=

❨

C

x

A

❩

=

❨

182

x

200

❩

= 182 : 169.

B

A

B

200

169

When C covers 182 m, B covers 169 m.

When C covers 350 m, B covers	❨	169	x 350	❩m	= 325 m.
		182

Therefore, C beats B by (350 - 325) m = 25 m.