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

Correct Option

Let the cost price (CP) of 1 unit of honey be Re. 1. Consequently, the CP of 100 units of honey is Rs. 100.

To achieve a 20% profit when selling the honey, the required selling price (SP) for these 100 units would be:

SP = CP + (Profit % of CP)

SP = 100 + (20% of 100) = 100 + 20 = Rs. 120.

Now, assume 'x' units of water are mixed with 100 units of honey. The total volume of the mixture becomes (100 + x) units.

The problem states that the mixture is sold at the cost price of honey, which is Re. 1 per unit. Therefore, the total selling price of the mixture is (100 + x) × Re. 1 = Rs. (100 + x).

To gain a 20% profit, the total selling price of the mixture must equal the required SP calculated earlier:

100 + x = 120

$\Rightarrow$⇒ x = 120 – 100 = 20 units.

Thus, 20 units of water are mixed with 100 units of honey. The percentage of water in the mixture, relative to the quantity of honey, is calculated as:

Percentage of water $= \frac{\text{Quantity of water}}{\text{Quantity of honey}} \times 100\%$=Quantity of honeyQuantity of water​×100%

$= \frac{20}{100} \times 100\% = 20\%$=10020​×100%=20%.

Incorrect Options

Options 2, 3, and 4 are incorrect. The calculation demonstrates that mixing 20% water with honey is necessary to achieve a 20% profit when the mixture is sold at the cost price of honey. These options do not satisfy the specified profit condition under the given parameters.