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
Initial ratio (acid : water) = 3 : 1
Mixture drawn off = 8 liters
New solution ratio = 1 : 3
Initial acid = (3/4) × 40 = 30 liters
Initial water = (1/4) × 40 = 10 liters
In 8 liters drawn off
Acid = (3/4) × 8 = 6 liters
Water = (1/4) × 8 = 2 liters
Remaining acid = 30 − 6 = 24 liters
Remaining water = 10 − 2 = 8 liters
In 8 liters added (1:3)
Acid = (1/4) × 8 = 2 liters
Water = (3/4) × 8 = 6 liters
Final acid = 24 + 2 = 26 liters
Final water = 8 + 6 = 14 liters
Final ratio = 26 : 14 = 13 : 7