Understanding Cost and Selling Price in Economics: A Real-World Application

In the world of economics and business, it is crucial to understand the relationship between cost price, selling price, profit, and loss. This article will delve into a real-world example of a scenario where a person, Gita, purchases two TV sets and sells them at different profit and loss percentages, with the selling price of both being the same.

Problem Statement

Gita purchased two TV sets for Rs. 30000. She then sold one TV set at a loss of 15% and the other at a gain of 19%, with the selling price of both being the same. The goal is to determine the cost price of each TV set.

Solving the Problem

We shall use algebra to solve this problem effectively. Let's denote the cost price of the first TV set as C_1 and the cost price of the second TV set as C_2.

According to the problem:

$C_1 C_2 30000$

The selling price of the first TV set, sold at a loss of 15%, can be calculated as:

$text{{Selling Price of TV 1}} C_1 - 0.15C_1 0.85C_1$

The selling price of the second TV set, sold at a gain of 19%, is:

$text{{Selling Price of TV 2}} C_2 0.19C_2 1.19C_2$

Since the selling prices of both TV sets are the same, we can set the equations equal to each other:

$0.85C_1 1.19C_2$

We now have two equations and can express C_1 in terms of C_2:

$C_1 frac{1.19C_2}{0.85}$

Substituting this expression for C_1 into the first equation:

$frac{1.19C_2}{0.85} C_2 30000$

Combining the terms and solving for C_2 involves some algebraic manipulation. Let's rewrite C_2 with a common denominator:

$frac{1.19C_2}{0.85} frac{0.85C_2}{0.85} 30000$

Combining the fractions:

$frac{1.19C_2 0.85C_2}{0.85} 30000$

This simplifies to:

$frac{2.04C_2}{0.85} 30000$

Multiplying both sides by 0.85:

$2.04C_2 30000 times 0.85$

Solving for C_2:

$C_2 frac{30000 times 0.85}{2.04} approx 12500$

Substituting C_2 back into the first equation to find C_1:

$C_1 30000 - C_2 30000 - 12500 17500$

We can conclude that the cost price of the first TV set is Rs. 17500 and the cost price of the second TV set is Rs. 12500.

Additional Examples

Let's consider a couple of additional scenarios to further illustrate the relationship between cost price, selling price, and profit or loss:

Example 1

If Gita had bought the first TV set at Rs. 23375 at a 10% loss and the second TV set at Rs. 19125 at a 10% profit, we can verify that the selling prices would be the same:

C.E.P of the 1st TV: Rs. 23375 at 10% loss:

$text{{Selling Price of 1st TV}} frac{23375 times 90}{100} 21037.5$

C.E.P of the 2nd TV: Rs. 19125 at 10% profit:

$text{{Selling Price of 2nd TV}} frac{19125 times 110}{100} 21037.5$

Both selling prices are indeed the same, confirming our earlier calculation.

Example 2

Let's consider a different scenario where the cost price of the first TV set is Rs. X and the second TV set is Rs. 30000 - X:

Selling price of the first TV set at 15% loss:

$text{{Selling Price of 1st TV}} X times frac{85}{100} 0.85X$

Selling price of the second TV set at 19% gain:

$text{{Selling Price of 2nd TV}} (3000 - X) times frac{119}{100} 3000 - X times 1.19$

Since the selling prices are the same:

$0.85X 3570 - 1.19X$

Combining like terms:

$2.04X 35700$

Solving for X:

$X frac{35700}{2.04} 17500$

The cost price of the first TV set is Rs. 17500, and the cost price of the second TV set is Rs. 12500 (30000 - 17500).

Conclusion

This article explored how to solve for the cost price of items sold at different profit and loss percentages using algebra. By understanding and applying the principles of cost price and selling price, businesses can make more informed decisions and optimize their financial performance.

Key Takeaways

Cost Price (CP): The original purchase price of an item. Selling Price (SP): The price at which an item is sold. Profit: Selling price minus cost price. Loss: Cost price minus selling price.

These concepts form the backbone of economic analysis and are crucial for both personal and business finance.