From a cask of wine containing 25 litres, 5 litres are withdrawn and the cask is refilled with water. The process is repeated a second and then a third time. Find the quantity of wine left in the cask and also the ratio of wine to water in the resulting mixture.

Average speed = (100 / 10) = 10 km/hr.
Ratio of time taken at 7 km/hr to 12 km/hr = 2 : 3
Time taken at 7 km/hr = 2 (2 + 3) × 10 = 4 hrs.
Distance covered at 7 km/hr = 7 × 4 = 28 km.
Distance covered at 12 km/hr = 100 ? 28 = 72 km.

First of all we write the fraction of milk present in three mixtures. In A : 7 /12
In B : 17 /24
In combination of A and B : 5 /8
We now apply allegation rule on these fractions from figure.
So, Ratio of A: B = 2 : 1


2nd Method
Let us assume P mixture taken from first vessel and Q mixture taken from second vessel to form a new mixture.
Part of Milk in P mixture from first vessel = 7P/12
Part of Milk in Q mixture from Second vessel = 17Q/24
Part of Water in P mixture from first vessel = 5P/12
Part of Milk in Q mixture from Second vessel = 7Q/24
According to question,
After mixing the P and Q, we will get mixture.
Milk in New Mixture / Water in New Mixture = 5/3
{(7P/12) + (17Q/24 ) } /{ (5P/12) + (7Q/24) } = 5/3
{(14P + 17Q)/24 } /{ (10P + (7Q)/24 } = 5/3
{(14P + 17Q) } /{ (10P + (7Q)} = 5/3
(14P + 17Q) x 3 = 5 x (10P + (7Q)
42P + 51Q = 50P + 35Q
51Q - 35Q = 50P - 42P
8P = 16Q
P = 2Q
P/Q = 2
P : Q = 2 : 1

Let us assume P liters of Diesel added to the mixture so that Diesel will be 25% in the new mixture.
According to Question,
Quantity of Diesel in 150 liters of the mixture = 20% of 150 = 150 x 20/100 = 30 liters.
After adding P liters of Diesel, The total quantity of Diesel becomes (30 + P) and total volume of mixture will be (150 + P).
Again According to Question,
After adding the P liters of Diesel in the mixture, the Diesel quantity becomes 25% of the new mixture.
Quantity of Diesel in new mixture = 25% of Total mixture.
? (30 + P) = (150 + P) x 25 %
? 30 + P = (150 + P) x 25/100
? 30 + P = (150 + P) x 1/4
? 120 + 4P = 150 + P
? 4P - P = 150 - 120
? 3P = 30
? P = 10 liters.

21.% of liquid B initially present in the vessel = 2 / (3 + 2) × 100 = 40%
% of liquid B finally present in the vessel = 4 / (1 + 4) × 100 = 80%
The second solution is liquid B which is being mixed and it has 100% liquid B.
80% of liquid B present in the resultant mixture may be taken as average percentage. So, using rule of alligation on liquid B per cent, we can write,
or 1 : 2
The ratio of liquid left in the vessel to liquid B being mixed = 1: 2
Since the quantity of liquid B being mixed is 20 liters, the quantity of liquid left in the vessel is 10 liters.
Therefore, the total quantity of liquid initially present in the vessel = 10 + 20 = 30 liters
Quantity of liquid A = 3 / (2 + 3) × 30 = 18 liters.

If the two alloys are mixed, the mixture would contain 15 gms of each metal and it would cost Rs. (150 + 120) = Rs. 270.
Cost of (15 gms of metal A + 15 gms of metal B) = Rs. 270
Cost of (1 gm of metal A + 1 gm of metal B) = Rs. (270 / 15) = Rs. 18
Cost of 1 gm of metal B = Rs. (18 ? 6) = Rs. 12
Average cost of original piece of alloy = (150 / 15) = Rs. 10 per gm.
Quantity of metal / A Quantity of metal B = (2 / 4) = (1 / 2)
Quantity of metal B = 2 (1 + 2) × 15 = 10 gms.

Here withdrawal of liquid A and B result into making the container empty. Hence percentage of two liquids withdrawn are two components of the percentage by which the container becomes empty.
Applying the rule of alligation, we get
A : B = 10 : 20 or 1 : 2
Quantity of liquid = 1 (1 + 2) × 90 = 30
liters Quantity of liquid B = 90 ? 30 = 60 liters.

Here first two varieties of tea are mixed in equal ratio.
So their average price = (126 + 135) /2 = Rs. 130.50
Let price of the third variety per kg be Rs. A; then now mixture is formed by two varieties one at Rs. 130.50 per kg and other at Rs. A per kg in the same ratio 2 : 2 i.e, 1 : 1
By the rule of alligation,
(A - 153) / (22.50) = 1
? A ? 153 = 22.50
? A = Rs. 175.50