Understanding the Honey and Water Mixture Problem

This problem involves a concept often seen in profit and loss, specifically dealing with mixtures and calculating the percentage of an added ingredient (water) when a certain profit margin is achieved by selling the mixture at the cost price of the main ingredient (honey).

The key information is:

• Water is mixed with honey.
• The mixture is sold at the cost price of the honey.
• A gain of 20% is made.

The gain comes solely from the water added, as water is assumed to have zero cost, while the selling price per unit of mixture is based on the cost price per unit of honey.

Step-by-Step Calculation

Let's assume a quantity of honey and its cost price to make the calculation straightforward.

1. Assume the cost price of 1 unit of honey is \(Rs. 1\).
2. Let's take a convenient quantity of honey, say 100 units. The total cost of this honey is \(100 \times Rs. 1 = Rs. 100\).
3. Water is added to this honey. Water has no cost, so the total cost of the mixture is still \(Rs. 100\).
4. The problem states that the mixture is sold at the cost price of honey. This implies that each unit of the mixture is sold at \(Rs. 1\) (the cost price per unit of honey).
5. A gain of 20% is achieved on the cost price of the honey.
6. Calculate the total gain: Gain \( = 20\% \text{ of Cost Price} = 0.20 \times Rs. 100 = Rs. 20\).
7. The total selling price of the mixture is the total cost plus the gain: Total SP \( = Rs. 100 + Rs. 20 = Rs. 120\).
8. The mixture is sold at \(Rs. 1\) per unit. If the total selling price is \(Rs. 120\), the total quantity of the mixture must be \( \frac{Rs. 120}{Rs. 1 \text{ per unit}} = 120 \text{ units} \).
9. The total quantity of the mixture is the sum of the quantity of honey and the quantity of water. Let \(Q_{honey}\) be the quantity of honey and \(Q_{water}\) be the quantity of water.
10. We started with \(Q_{honey} = 100\) units. The total mixture quantity is \(Q_{mixture} = 120\) units.
11. The quantity of water added is \(Q_{water} = Q_{mixture} - Q_{honey} = 120 \text{ units} - 100 \text{ units} = 20 \text{ units}\).
12. The question asks for the percentage of water mixed with honey. This means \(\frac{\text{Quantity of Water}}{\text{Quantity of Honey}} \times 100\%\).
13. Percentage of water \( = \left(\frac{20 \text{ units}}{100 \text{ units}}\right) \times 100\% = 0.20 \times 100\% = 20\%\).

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

Summary of Results

The percentage of water mixed with honey is \(\frac{\text{Quantity of Water}}{\text{Quantity of Honey}} \times 100\% = \frac{20}{100} \times 100\% = 20\%\).

Revision Table: Key Concepts

Additional Information: Zero Cost Ingredient

In problems involving mixtures where one ingredient (like water) has zero cost, any profit made when selling the mixture at a price related to the costly ingredient comes from the quantity of the zero-cost ingredient added. When the mixture is sold at the cost price of the original costly ingredient, the total selling price of the mixture will be greater than the cost of the original costly ingredient. The difference is the profit, and this profit is effectively 'earned' on the added quantity of the zero-cost ingredient.

The relationship between the quantity of the costly ingredient (\(Q_C\)) and the zero-cost ingredient (\(Q_Z\)) for a given profit percentage (P%) when selling the mixture at the cost price of the costly ingredient is often simplified to the ratio:

In this honey and water problem:

• \(Q_Z\) is the quantity of water.
• \(Q_C\) is the quantity of honey.
• \(P\) is the profit percentage (20%).

So, \( \frac{Q_{water}}{Q_{honey}} = \frac{20}{100} = \frac{1}{5} \). This means for every 5 parts of honey, 1 part of water is added.

The percentage of water mixed with honey is \(\left(\frac{Q_{water}}{Q_{honey}}\right) \times 100\% = \frac{1}{5} \times 100\% = 20\%\).

This confirms the result obtained through the step-by-step calculation.