What percent of water must be mixed with honey so as to gain 20% by selling the mixture at the cost price of honey?

The correct option is 20%.

Let the cost price of 1 unit volume of pure honey be $C$.

Suppose we have $H$ units of honey and we mix it with $W$ units of water.

Since water is assumed to be free (cost is 0), the total cost price of the mixture is only the cost of the honey:
Total Cost Price (CP) = $H\times C$

The mixture is sold at the cost price of pure honey, which is $C$ per unit volume. The total volume of the mixture is $H+W$.
Total Selling Price (SP) = $(H+W)\times C$

The profit gained is the difference between the Selling Price and the Cost Price:
Profit = SP - CP = $(H+W)C-HC=W\times C$

We are given that the profit percentage is 20%. The formula for profit percentage is:
Profit % = $\frac{\text{Profit}}{\text{Cost Price}}\times 100$

Substituting the values into the formula:
$20=\frac{W\times C}{H\times C}\times 100$

Canceling $C$ from both the numerator and the denominator, we get:
$20=\frac{W}{H}\times 100$

Solving for the ratio of water to honey:
$\frac{W}{H}=\frac{20}{100}=\frac{1}{5}$

To find what percent of water must be mixed with honey (i.e., the volume of water as a percentage of the volume of honey):
Percentage of water mixed with honey = $\frac{W}{H}\times 100=\frac{1}{5}\times 100=20\%$

Thus, 20% of water must be mixed with the honey to achieve a 20% gain when selling the mixture at the cost price of honey.