A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

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.

S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.

∴ C.P. of 1 kg of mixture = Rs.	❨	100	x 9.24	❩	= Rs. 8.40
		110

By the rule of allilation, we have:

C.P. of 1 kg sugar of 1st kind Cost of 1 kg sugar of 2nd kind
Rs. 9	Mean Price Rs. 8.40	Rs. 7
1.40		0.60

∴ Ratio of quantities of 1^st and 2^nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

Then, 7 : 3 = x : 27

⟹ x =	❨	7 x 27	❩	= 63 kg.
		3

By the rule of alligation:

Cost of 1 kg of 1st kind Cost of 1 kg of 2nd kind
720 p	Mean Price 630 p	570 p
60		90

∴ Required ratio = 60 : 90 = 2 : 3.

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

Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.

C.P. of 1 litre mixture = Re.	❨	100	x 1	❩	=	4
		125				5

By the rule of alligation, we have:

C.P. of 1 litre of milk C.P. of 1 litre of water
Re. 1	Mean Price Re. 4 5	0
4 5		1 5

∴ Ratio of milk to water =	4	:	1	= 4 : 1.
	5		5

Hence, percentage of water in the mixture =	❨	1	x 100	❩%	= 20%.
		5

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 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 the quantity of the wine in the cask originally be x litres.

Then, quantity of wine left in cask after 4 operations =	[	x	❨	1 -	8	❩	4	] litres.
					x

∴	❨	x(1 - (8/x))4	❩	=	16
		x			81

⟹	❨	1 -	8	❩	4	=	❨	2	❩	4
			x					3

⟹	❨	x - 8	❩	=	2
		x			3

⟹ 3x - 24 = 2x

⟹ x = 24.

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.