140 liters of a solution is formed by mixing two solutions A and B. The concentrations of milk in A and B are 40% and 75%, respectively. If the resultant solution has a milk concentration of 52%, then what is the quantity of A that is there in the resultant solution?

The concentrations of milk in A and B are 40% and 75%, respectively.

Ratio of milk and remaining part in solution A = 40% : (100-40)%

= 40% : 60%

= 2 : 3

So let's assume the milk and remaining part in solution A is 2y and 3y.

Ratio of milk and remaining part in solution B = 75% : (100-75)%

= 75% : 25%

= 3 : 1

So let's assume the milk and remaining part in solution B is 3z and z.

If the resultant solution has a milk concentration of 52%.

2y+3z = 140 of 52%

2y+3z = 72.8    Eq.(i)

3y+z = 140 of (100-52)%

3y+z = 140 of 48%

3y+z = 67.2

Multiply the above equation by 3.

9y+3z = 201.6    Eq.(ii)

Eq.(ii) - Eq.(i)

9y+3z-(2y+3z) = 201.6-72.8

9y+3z-2y-3z = 201.6-72.8

7y = 128.8

y = 18.4

Quantity of A in the resultant solution = 2y+3y

= 5y

= $$5\times18.4$$

= 92 litres