A mixture of 15 liters of milk and water contains 5% water. How much pure milk should be added to get a mixture containing 2% water ?

Correct option is D

Given:

Total mixture = 15 liters
Water percentage = 5%

Formula Used:

​$$\text{Final water percentage} = \frac{\text{Water}}{\text{Milk + Water}} \times 100$$​​

Solution:

Initial mixture = 15 liters
Water content = 5% of 15 = 0.75 liters
Milk content = 15 - 0.75 = 14.25 liters
Let x liters of pure milk be added.

​$$\frac{0.75}{15 + x} \times 100 = 2$$

$$\frac{0.75}{15 + x} = 0.02$$

​​​0.75 = 0.02(15 + x)

0.75 = 0.3 + 0.02x

0.45 = 0.02x

​$$x = \frac{0.45}{0.02} = 22.5$$ liters

So, 22.5 liters of pure milk should be added.​​

Thus, the correct answer is (d).

Alternative Method:

By ratio:

20x = 15

x = $$\frac{15}{20} = \frac{3}{4}$$

Add milk = 30x = $$30 \times \frac{3}{4} = \frac{90}{4} = 22.5$$ liters​