A mixture of 40L of alcohol and water contains 10% water. How much water should be added to this mixture, so that the new mixture contains 20% water?

The correct option is 5L.

Let's solve the problem step-by-step to find out how much water needs to be added to make the water concentration 20% in the new mixture.

Step 1: Find the initial quantity of water and alcohol in the 40L mixture.

Total volume of the initial mixture = 40 L
Percentage of water = 10%

Initial quantity of water:

$\text{Water}=10\%\text{ of }40=\frac{10}{100}\times 40=4\text{ L}$

Initial quantity of alcohol:

$\text{Alcohol}=40-4=36\text{ L}$

Step 2: Set up the equation for adding water.

Let the amount of water to be added be $x$ liters.

When $x$ liters of water is added:
- New quantity of water = $(4+x)$ L
- New total volume of the mixture = $(40+x)$ L

We are given that the new mixture should contain 20% water. Therefore:

$\frac{4+x}{40+x}=20\%=\frac{20}{100}=\frac{1}{5}$

Step 3: Solve for $x$.

Cross-multiplying the equation:

$5\times (4+x)=1\times (40+x)$

$20+5x=40+x$

Subtracting $x$ from both sides:

$4x+20=40$

Subtracting 20 from both sides:

$4x=20$

$x=\frac{20}{4}=5$

Thus, 5L of water should be added to the mixture.