Question

a milkman mixed two solutions A and B of milk and water in the ratio of 4:5 respectively, then the new solution obtained in which the concentration of milk was 75%. in a 5 litre solution A, the quantity of water was 2 litre, then in 15 litre solution B , what was the quantity (in ml) of milk?​

Answer 1

Given :-

• A milkman mixed two solutions A and B of milk and water in the ratio of 4:5 respectively.
• A new solution was obtained in which the concentration of milk was 75%.
• In 5L solution of A, the quantity of water was 2L.

To Find :-

• In 15L solution of B, the quantity of milk ?

Solution :-

Let us Assume that, Quantity of Milk in Solution B of 15 litre is x Litre.

so,

→ water in A = 2 litre .

→ milk in A = 5 - 2 = 3 litre.

Than,

→ (Milk in A) / (water in A) = (3/2)

and,

→ milk in B = x litre.

→ water in B = (15 - x) Litre.

Than,

→ (Milk in B) / (water in B) = x/(15 - x)

Also,

→ in final concentration milk = 75%

→ water = 100% - 75% = 25% .

Than,

→ (Milk in final concentration) / (water in final concentration) = 75/25 = (3/1).

Also,

→ (Milk in A + water in A) / (Milk in B + water in B) = 4/5.

______________

Now,

→ (Milk in final concentration) / (water in final concentration) = 75/25 = (3/1).

→ (Milk in final concentration) = 3*(water in final

concentration)

→ Total milk of (A + B) = 3 * Total water of (A + B)

→ {(3/5) + (x / 15)} = 3 * [(2/5) + {(15 - x)/15}]

→ {(3/5) + (x / 15)} = (6/5) + {3(15 - x)/15}

→ {(3/5) + (x / 15)} = (6/5) + (15 - x)/5

→ {(3/5) + (x / 15)} = { (6 + 15 - x) / 5 }

→ (9 + x) / 15 = (21 - x) / 5

→ (9 + x) = 3(21 - x)

→ 9 + x = 63 - 3x

→ x + 3x = 63 - 9

→ 4x = 54

→ x = 13.5 Litre. (Ans.)

Hence, Quantity of milk in Solution B was 13.5 Litre.