A 50-litre mixture of oil and water has 16% water. How many litres of oil must be added to make water content 10%?

The correct answer is 30 litres.

Let's solve this problem step-by-step by calculating the quantity of water and oil present in the mixture.

Step 1: Calculate the initial quantities of water and oil in the mixture.

Total volume of the initial mixture = 50 litres.

Percentage of water in the initial mixture = 16%.

Quantity of water = 16% of 50 litres

$\text{Quantity of water}=\frac{16}{100}\times 50=8\text{ litres}$

Quantity of oil in the initial mixture = Total volume - Water volume

$\text{Quantity of oil}=50-8=42\text{ litres}$

Step 2: Determine the new mixture conditions.

We are adding pure oil to the mixture, which means the quantity of water remains constant at 8 litres.

In the new mixture, water should make up 10% of the total volume.

Step 3: Calculate the total volume of the new mixture.

Let the total volume of the new mixture be $V$ litres.

Since 10% of the new mixture is water:

$10\%\text{ of }V=8\text{ litres}$

$\frac{10}{100}\times V=8$

$V=8\times 10=80\text{ litres}$

Step 4: Find the amount of oil added.

Oil added = New total volume - Initial total volume

$\text{Oil added}=80-50=30\text{ litres}$

Therefore, 30 litres of oil must be added to make the water content 10%.