Understanding the Profit and Loss Problem

This problem involves a sequence of sales, where goods are sold from X to Y, and then from Y to Z, with profits added at each step. We are given the final cost price for Z and the profit percentages at each stage. Our goal is to find the initial cost price for X.

Step-by-Step Calculation of Cost Price for X

Let's break down the problem and calculate the cost price at each step:

1. Cost Price for X (\(C_X\)): This is what we need to find. Let's represent it as \(C_X\).
2. Selling Price by X to Y (\(S_X\)): X sells the goods to Y at a profit of 20 percent. The selling price for X is the cost price for Y (\(C_Y\)).
   The profit is 20% of \(C_X\).
   Profit = \(0.20 \times C_X\)
   Selling Price \(S_X\) = Cost Price \(C_X\) + Profit
   \(S_X = C_X + 0.20 C_X = (1 + 0.20) C_X = 1.20 C_X\)
   So, the cost price for Y is \(C_Y = S_X = 1.20 C_X\).
3. Selling Price by Y to Z (\(S_Y\)): Y sells the same goods to Z at a profit of 50 percent. The selling price for Y is the cost price for Z (\(C_Z\)).
   The profit is 50% of \(C_Y\).
   Profit = \(0.50 \times C_Y\)
   Selling Price \(S_Y\) = Cost Price \(C_Y\) + Profit
   \(S_Y = C_Y + 0.50 C_Y = (1 + 0.50) C_Y = 1.50 C_Y\)
   So, the cost price for Z is \(C_Z = S_Y = 1.50 C_Y\).

Relating the Cost Prices Using Profit Percentages

We have two relationships:

• \(C_Y = 1.20 C_X\)
• \(C_Z = 1.50 C_Y\)

We can substitute the first equation into the second one to relate \(C_Z\) directly to \(C_X\):

\(C_Z = 1.50 \times (1.20 C_X)\)

\(C_Z = (1.50 \times 1.20) C_X\)

\(C_Z = 1.80 C_X\)

Finding the Initial Cost Price for X

We are given that the cost price for Z is Rs 9000. So, \(C_Z = 9000\).

Now, we can use the relationship \(C_Z = 1.80 C_X\) to find \(C_X\):

\(9000 = 1.80 C_X\)

To find \(C_X\), we need to divide 9000 by 1.80:

\(C_X = \frac{9000}{1.80}\)

\(C_X = \frac{9000}{\frac{18}{10}}\)

\(C_X = \frac{9000 \times 10}{18}\)

\(C_X = \frac{90000}{18}\)

\(C_X = 5000\)

Therefore, the cost of the goods for X is Rs 5000.

Substituting \(C_Y\) into the second relationship:

\(C_Z = 1.50 \times (1.20 C_X)\)

\(C_Z = 1.80 C_X\)

Given \(C_Z = 9000\), we solve for \(C_X\):

\(9000 = 1.80 C_X\)

\(C_X = \frac{9000}{1.8} = \frac{90000}{18} = 5000\)

Revision Table: Key Concepts in Profit and Loss

Additional Information on Sequential Sales and Profit Calculation

In problems involving sequential sales like this, the selling price of the first seller becomes the cost price for the second buyer, and so on. When profit is calculated as a percentage, it is usually based on the cost price for the person who is selling.

If a profit of P% is made on a cost price CP, the selling price SP is calculated as:

\(SP = CP + \left(\frac{P}{100} \times CP\right) = CP \left(1 + \frac{P}{100}\right)\)

In our problem:

• Y buys from X at \(C_Y\), which is X's selling price (\(S_X\)). X made a 20% profit on \(C_X\).
  \(C_Y = S_X = C_X \left(1 + \frac{20}{100}\right) = C_X (1 + 0.20) = 1.20 C_X\)
• Z buys from Y at \(C_Z\), which is Y's selling price (\(S_Y\)). Y made a 50% profit on \(C_Y\).
  \(C_Z = S_Y = C_Y \left(1 + \frac{50}{100}\right) = C_Y (1 + 0.50) = 1.50 C_Y\)

Combining these relationships allows us to link the initial cost \(C_X\) to the final cost \(C_Z\).