Bottle A contains 100 ml of 20% saline solution, while bottle B contains 40 ml of 40% saline solution. 10 ml of water is accidently added to the bottle B. What amount of solution from bottle A should be poured in Bottle B to obtain a 25% saline solution?

Bottle A contains 100 ml of 20% saline solution, while bottle B contains 40 ml of 40% saline solution. 10 ml of water is accidently added to the bottle B. What amount of solution from bottle A should be poured in Bottle B to obtain a 25% saline solution? Correct Answer 70 ml

GIVEN :

Bottle A contains 100 ml of 20% saline solution, while bottle B contains 40 ml of 40% saline solution.

10 ml of water is accidently added to the bottle B.

CONCEPT :

Addition of new liquid.

ASSUMPTION :

Let the amount of solution poured from bottle A be ‘x’ ml

CALCULATION :

Initial amount of salt in 40 ml solution of bottle B = 40% of 40 = 16 ml

After addition of 10 m water, amount of salt remains the same, but,

Percentage of salt in (40 + 10) = 50 ml solution of bottle B = (16/50) × 100 = 32%

Hence, the final solution will contain,

⇒ 20% of x + 32% of 50 = 25% of (x + 50)

⇒ 0.2x + 16 = 0.25x + 12.5

⇒ 0.25x – 0.2x = 16 – 12.5

⇒ 0.05x = 3.5

⇒ x = 3.5/0.05 = 70 ml