A shopkeeper buys 26 liter of milk @ Rs. 16 per liter. He also buys from another source an inferior quality of milk @ Rs. 10 per liter. How much quantity of the later should he buy to mix it with the former so that he can sell the mixture @ Rs. 14 per liter without making any loss?

Let us assume the ratio product is Q.
According to question,
Quantity of Milk in mixture = 6Q ml
Quantity of Water in mixture = 2Q ml
Quantity of Milk in mixture + Quantity of Water in mixture = 640
6Q + 2Q = 640
8Q = 640
Q = 640/8
Q = 80
Quantity of Milk in mixture = 6Q ml = 6 x 80 = 480 ml
Quantity of Water in mixture = 2Q ml = 2 x 80 = 160 ml
To find the new mixture containing half milk and half water,
Quantity of water to be added for making the mixture of equal quantity of milk and water = 480 - 160 = 320

According to question,
Quantity of Milk/Quantity of Water = 3/2
Let us assume the product ratio = n.
Quantity of Milk = 3n and Quantity of Water = 2n liters
Quantity of Milk + Quantity of Water = 20 liters
3n + 2n = 20 liters
5n = 20 liters
n = 20/5
n = 4
Quantity of Milk = 3n liters
Put the value of n,
Quantity of Milk = 3 x 4 = 12 liters
Quantity of Water = 2n
Quantity of Water = 2 x 4 = 8 liters
If 10 liters of mixture are removed first time, we will find how much milk and water contain in mixture.
? 20 liters of mixture contains 12 liter of milk.
? 1 liters of mixture contains 12/20 liter of milk.
? 10 liters of mixture contains 10 x 12/20 liter of milk.
? 10 liters of mixture contains 6 liter of milk.
? 10 liters of mixture contains 4 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters and water = 4 liters.
If the process is repeated one more time and 10 liters of the mixture are removed second time, then
If 10 liters of mixture are removed, Again we will find how much milk and water contain in the mixture.
? 20 liters of mixture contains 16 liter of milk.
? 1 liters of mixture contains 16/20 liter of milk.
? 10 liters of mixture contains 10 x 16/20 liter of milk.
? 10 liters of mixture contains 8 liter of milk.
? 10 liters of mixture contains 2 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 8 liters and amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 - 2) = 2 liters.
Now 10 liters milk is added => total milk = 18 liters and water will be 2 liters.
The required ratio of milk and water in the final mixture obtained
Quantity of milk/Quantity of Water= 18:2 = 9:1

The given solution has 75% milk.
Milk to be added has 100% milk.
Milk should be added to the given mixture in the ratio 15 : 10 or 3 : 2
? Quantity of milk to be added = (3 / 2) × 6 = 9 liters.

Let cost price of spirit be Re. 1 per liter.
Then SP of mixture = Re. 1 per liter
Gain = 25%
So, CP of mixture = 1 × (100 / 125) = Re. 4 / 5
We assume that CP of water is zero.
Using allegation rule on cost price,
Water should be mixed to spirit in the ratio (1 / 5) : (4 / 5) or 1 : 4

According to figure we find that the ratio will be 3 : 1.

Quantity sold at 20% profit = 3 / (3 + 1) × 50 = 37.5 kgs.
Quantity sold at 40% profit = (50 ? 37.5) = 12.5 kgs.

As per figure we can calculate the ration as below
Number of officers / Number of workers = 1000 / 7000 = 1 / 7
No. of officers = 1 / (1 + 7) × 400 = 50
No. of workers = 400 ? 50 = 350