A container of 90 liters with two liquids A and B, 60% of liquid A and 30% of liquid B are taken out of the vessel. This leaves the container 40% empty. Find the initial quantity of both liquids.

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.

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.

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

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