From a 50-litre mixture containing alcohol and water in the ratio 3:2, 10 litres are removed and replaced with water.
What is the new ratio of alcohol to water?

Correct option is D

Given:
Initial volume = 50 L
Initial ratio of alcohol to water = 3:2
Volume of mixture removed = 10 L
Volume of water added = 10 L
Formula Used:
Quantity of component = Total volume × Fraction of component
Solution:
Initial volume of alcohol = 50 × $$\frac{3}{5}$$​ = 30 L
Initial volume of water = 50 ×$$\frac{2}{5}$$​= 20 L
Volume of alcohol in 10 L removed = 10 × $$\frac{3}{5}$$​ = 6 L
Volume of water in 10 L removed = 10 × $$\frac{2}{5}$$​ = 4 L
Remaining volume of alcohol = 30 - 6 = 24 L
Remaining volume of water = 20 - 4 = 16 L
Now, 10 L of water is added.
New volume of water = 16 + 10 = 26 L
New ratio of alcohol to water = 24:26 = 12:13
Final Answer
So the correct answer is (d)