Question

A 100-litre solution contains acid and water in the ratio 3:2. Some quantity of this solution is removed and replaced with pure acid. If the final ratio of acid to water becomes 7:3, how many litres of solution were replaced?

• A. 35 litres
• B. 25 litres
• C. 50 litres
• D. 20 litres

Answer 1

Answer:

B. 25 litres

Initial solution = 100 litres

Ratio of acid : water = (3:2)

So,

• Acid $(= \frac{3}{5}\times100 = 60) L$

• Water $(= \frac{2}{5}\times100 = 40) L$

Let (x) litres of solution be removed.

Since the solution is removed in the same ratio:

• Acid removed $(= \frac{3x}{5})$

• Water removed $(= \frac{2x}{5})$

After removal:

• Acid left $(= 60 - \frac{3x}{5})$

• Water left $(= 40 - \frac{2x}{5})$

Now (x) litres of pure acid are added.

Final acid:

• $60-\frac{3x}{5}+x = 60+\frac{2x}{5}$
  

Final water:

• $40-\frac{2x}{5}$
  ]

Given final ratio (= 7:3):

• $\frac{60+\frac{2x}{5}}{40-\frac{2x}{5}}=\frac{7}{3}$

Cross multiply:

$3\left(60+\frac{2x}{5}\right)=7\left(40-\frac{2x}{5}\right)$
$180+\frac{6x}{5}=280-\frac{14x}{5}$
$\frac{20x}{5}=100$
$4x=100$
$x=25$

Therefore, 25 litres of solution were replaced.