A vessel contains mixture of liquids A and B in the ratio 3: 2. When 20 litres of the mixture is taken out and replaced by 20 litres of liquid B, the ratio changes to 1: 4. How many litres of liquid A was there initially present in the vessel?

Here, quantity of wine left after third operation
= [1 - (5 / 25)]³ x 25 = (4 / 5)³ x 25 = (64 / 125) x 25 = (64 / 5) = 12 ⁴/₅ liters.
Final ratio of wine to water = (64 / 125) / (1- 64 /125)
= (64 / 125) /(61 / 125)
Wine : Water = (64 / 61)

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.

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