A 96-litre solution contains acid and water in the ratio 5:3. Some quantity of this solution is removed and replaced with pure acid. If the final ratio of acid to water becomes 11:5, how many litres of solution were replaced?
Show answer & explanation
Correct answer: B
Explanation
Answer: 16 litres
Working:
Total solution = 96 litres, acid:water = 5:3
Acid = 60 litres, Water = 36 litres
Let x litres of solution be removed (containing acid and water in same 5:3 ratio), then replaced with pure acid.
Removed: acid = (5/8)x, water = (3/8)x
After removing and adding x litres pure acid:
Acid = 60 - (5/8)x + x = 60 + (3/8)x
Water = 36 - (3/8)x
New ratio = 11:5
[60+(3/8)x] / [36-(3/8)x] = 11/5
5(60+(3/8)x) = 11(36-(3/8)x)
300 + (15/8)x = 396 - (33/8)x
(48/8)x = 96
6x = 96
x = 16
Verification:
Acid = 60 + (3/8)(16) = 60 + 6 = 66
Water = 36 - (3/8)(16) = 36 - 6 = 30
Ratio = 66:30 = 11:5 ✓
x = 16 litres