Expenses), Which Were 9000, Calculate His 7. After Spending *75,000 For Remodelling A House, Sania Sold The House At A Profit Price Was 36,00,000, How Much Did She Spend To Buy The House?
Introduction
In this article, we will delve into a mathematical problem involving Sania, who spent a significant amount on remodelling a house and then sold it at a profit. We will use algebraic equations to solve for the unknowns and find out how much Sania spent to buy the house.
Problem Statement
Sania spent *75,000 on remodelling a house. After the remodelling, she sold the house at a profit of *36,00,000. We need to find out how much she spent to buy the house.
Step 1: Define the Variables
Let's define the variables:
- x: The cost price of the house (in rupees)
- y: The profit made by Sania (in rupees)
Step 2: Write the Equation
We know that Sania spent *75,000 on remodelling the house. This means that the cost price of the house plus the remodelling cost is equal to the selling price. We can write this as an equation:
x + 75,000 = 36,00,000 + y
Step 3: Simplify the Equation
We can simplify the equation by subtracting 75,000 from both sides:
x = 36,00,000 + y - 75,000
x = 35,25,000 + y
Step 4: Find the Value of y
We know that Sania made a profit of *36,00,000. This means that the selling price minus the cost price is equal to the profit. We can write this as an equation:
36,00,000 = x + y
Substituting the value of x from Step 3, we get:
36,00,000 = 35,25,000 + y + y
36,00,000 = 35,25,000 + 2y
Step 5: Solve for y
We can solve for y by subtracting 35,25,000 from both sides:
36,00,000 - 35,25,000 = 2y
10,75,000 = 2y
y = 10,75,000 / 2
y = 5,37,500
Step 6: Find the Value of x
Now that we have the value of y, we can find the value of x by substituting y into the equation from Step 3:
x = 35,25,000 + 5,37,500
x = 40,62,500
Conclusion
In this article, we solved a mathematical problem involving Sania, who spent a significant amount on remodelling a house and then sold it at a profit. We used algebraic equations to find out how much she spent to buy the house. The final answer is that Sania spent *40,62,500 to buy the house.
Key Takeaways
- We defined the variables x and y to represent the cost price and profit made by Sania.
- We wrote an equation to represent the relationship between the cost price, remodelling cost, and selling price.
- We simplified the equation and found the value of y.
- We used the value of y to find the value of x.
Real-World Applications
This problem can be applied to real-world scenarios where individuals or businesses need to calculate the cost of a project or the profit made from a sale. By using algebraic equations, we can solve for the unknowns and make informed decisions.
Future Directions
Introduction
In our previous article, we solved a mathematical problem involving Sania, who spent a significant amount on remodelling a house and then sold it at a profit. We used algebraic equations to find out how much she spent to buy the house. In this article, we will provide a Q&A section to clarify any doubts and provide additional information.
Q: What is the cost price of the house?
A: The cost price of the house is *40,62,500.
Q: How much did Sania spend on remodelling the house?
A: Sania spent *75,000 on remodelling the house.
Q: What is the profit made by Sania?
A: Sania made a profit of *36,00,000.
Q: How did you find the value of y?
A: We found the value of y by subtracting 35,25,000 from both sides of the equation 36,00,000 = 35,25,000 + 2y.
Q: What is the relationship between the cost price, remodelling cost, and selling price?
A: The relationship between the cost price, remodelling cost, and selling price is represented by the equation x + 75,000 = 36,00,000 + y.
Q: Can you explain the concept of algebraic equations?
A: Algebraic equations are mathematical statements that contain variables and constants. They are used to represent relationships between variables and can be solved to find the value of the variables.
Q: How can algebraic equations be applied to real-world scenarios?
A: Algebraic equations can be applied to real-world scenarios where individuals or businesses need to calculate the cost of a project or the profit made from a sale. By using algebraic equations, we can solve for the unknowns and make informed decisions.
Q: What are some common applications of algebraic equations?
A: Some common applications of algebraic equations include:
- Calculating the cost of a project
- Finding the profit made from a sale
- Determining the interest rate on a loan
- Calculating the time it takes to complete a task
Q: Can you provide more examples of algebraic equations?
A: Here are a few more examples of algebraic equations:
- 2x + 5 = 11
- x - 3 = 7
- 4x = 24
Conclusion
In this article, we provided a Q&A section to clarify any doubts and provide additional information on Sania's remodelling and selling puzzle. We also discussed the concept of algebraic equations and their applications in real-world scenarios.
Key Takeaways
- Algebraic equations are mathematical statements that contain variables and constants.
- They are used to represent relationships between variables and can be solved to find the value of the variables.
- Algebraic equations can be applied to real-world scenarios where individuals or businesses need to calculate the cost of a project or the profit made from a sale.
Real-World Applications
Algebraic equations have numerous real-world applications, including:
- Calculating the cost of a project
- Finding the profit made from a sale
- Determining the interest rate on a loan
- Calculating the time it takes to complete a task
Future Directions
In future articles, we can explore more complex mathematical problems involving algebraic equations. We can also apply these concepts to real-world scenarios and provide practical solutions.