A 40-litres mixture of acid and water contains acid and water in the ratio 3:1. 8 litres of this mixture is removed and replaced with another solution of acid and water in the ratio 1:3. What is the final ratio of acid to water in the resulting mixture?
Show answer & explanation
Correct answer: B
Explanation
Given:
Total mixture = 40 liters, Ratio (Acid : Water) = 3 : 1
According to the formula:
Part = (Ratio / Total Parts) × Total Quantity
Total parts = 3 + 1 = 4
Acid = (3/4) × 40 = 30
Water = (1/4) × 40 = 10
8 liters of the mixture was withdrawn.
Acid withdrawn = (3/4) × 8 = 6
Water withdrawn = (1/4) × 8 = 2
Remaining acid = 30 − 6 = 24
Remaining water = 10 − 2 = 8
Now, 8 liters of a solution with a ratio of (1:3) was added.
Acid = (1/4) × 8 = 2
Water = (3/4) × 8 = 6
New acid = 24 + 2 = 26
New water = 8 + 6 = 14
Ratio = 26 : 14 = 13 : 7