Question

A solution contains 20% of salt, from which 50 kg of water is removed. The quantity of salt remains 40% in that solution. Find the initial quantity of solution.

Answer 1

Let initial quantity of solution is x

so, initial amount of salt in solution = 20% of x = 20x/100 = x/5

and amount of water in solution = x - x/5 = 4x/5

a/c to question, 50kg of water is removed form solution.

e.g., amount of water remaining in solution = 4x/5 - 50

now new quantity of salt remains 40% in that solution.

so, percentage quantity of solution = {quantity of solution/amount of solution(salt + water)} × 100

40 % = (x/5)/(x/5 + 4x/5 - 50) × 100

or, 40 = (x/5)/(x - 50) × 100

or, 2/5 = x/5(x - 50)

or, 2 = x/(x - 50)

or, 2x - 100 = x

or, x = 100

hence, amount of solution is 100kg

Answer 2

Answer:

100 Kg

Explanation:

Concentration of the given solution is 20% which means in 100 kg of solution, 20 kg is salt and 80 kg is water.

Let the total mass of solution be x kg

Mass of salt = 0.20 x

Mass of water = 0.80 x

When 50 kg of water is removed, the amount of water in the solution changes however amount of salt in the solution remains same as earlier.

Now,

Mass of salt = 0.20 x

Mass of water = (0.80 x - 50)

Total mass of solution = 0.20 x + 0.80 x - 50

                                     = (x - 50) kg

According to question, concentration of salt is = 40%

(Mass of salt/ Mass of solution) x 100 = 40

⇒0.20 x/(x - 50) = 0.40

⇒0.20 x = 0.40 x - 20

⇒0.20 x = 20

⇒x = 100

Thus, initial quantity of solution = 100 Kg