Product A is costlier than product B by Rs. 2. If the price of product A is increased by two times the price of product B. the new price of product A become Rs. 17. What is the price of product B?
Rs. 5
Solving Product Price Calculations
This problem involves setting up algebraic equations to represent the given information about the prices of two products, Product A and Product B, and then solving for the unknown price of Product B.
Setting Up the Equations for Product Prices
Let's denote the price of Product A as \(P_A\) and the price of Product B as \(P_B\). We are given two pieces of information that can be translated into equations.
- Product A is costlier than product B by Rs. 2.
This means the price of Product A is equal to the price of Product B plus Rs. 2.
In equation form:
\(P_A = P_B + 2\) (Equation 1)
- The price of product A is increased by two times the price of product B, and the new price becomes Rs. 17.
The increase amount is two times the price of Product B, which is \(2 \times P_B\).
The new price of Product A is its original price plus this increase: \(P_A + 2 \times P_B\).
We are told this new price is Rs. 17.
In equation form:
\(P_A + 2P_B = 17\) (Equation 2)
Solving the Price Equations
We now have a system of two linear equations with two variables \(P_A\) and \(P_B\):
Equation 1: \(P_A = P_B + 2\)
Equation 2: \(P_A + 2P_B = 17\)
We can solve this system using the substitution method. Since Equation 1 already gives us an expression for \(P_A\) in terms of \(P_B\), we can substitute this expression into Equation 2.
Substitute \(P_B + 2\) for \(P_A\) in Equation 2:
\((P_B + 2) + 2P_B = 17\)
Now, we simplify and solve for \(P_B\):
\(P_B + 2 + 2P_B = 17\)
Combine like terms:
\(3P_B + 2 = 17\)
Subtract 2 from both sides of the equation:
\(3P_B = 17 - 2\)
\(3P_B = 15\)
Divide both sides by 3 to find \(P_B\):
\(P_B = \frac{15}{3}\)
\(P_B = 5\)
So, the price of Product B is Rs. 5.
Verifying the Product Prices
Let's check if our value for \(P_B\) satisfies the original conditions. If \(P_B = 5\), then from Equation 1:
\(P_A = P_B + 2\)
\(P_A = 5 + 2\)
\(P_A = 7\)
So, the original price of Product A is Rs. 7.
Now, let's check the second condition. The price of Product A is increased by two times the price of Product B. Two times the price of Product B is \(2 \times 5 = 10\).
The new price of Product A is its original price plus this increase:
New \(P_A = 7 + 10 = 17\)
This matches the information given in the question that the new price of Product A becomes Rs. 17.
Therefore, our calculated price for Product B (Rs. 5) is correct.
Conclusion on Product Pricing
The problem required setting up and solving a simple system of linear equations derived from the given information about the prices of Product A and Product B. By using substitution, we found the value of the variable representing the price of Product B.
| Variable | Represents | Calculated Value |
|---|---|---|
| \(P_A\) | Original Price of Product A | Rs. 7 |
| \(P_B\) | Original Price of Product B | Rs. 5 |
Revision Table: Understanding Algebraic Word Problems
| Concept | Explanation | Example Application |
|---|---|---|
| Identifying Variables | Assigning letters (like \(x\), \(y\), or \(P_A\), \(P_B\)) to unknown quantities in the problem. | Let \(P_B\) be the price of product B. |
| Translating Words to Equations | Converting phrases like "costlier than by" or "increased by" into mathematical operations and equations. | "A is costlier than B by 2" becomes \(P_A = P_B + 2\). |
| Solving Systems of Equations | Using methods like substitution or elimination to find the values of multiple variables from multiple equations. | Substitute \(P_A\) from one equation into another. |
| Verification | Plugging the calculated values back into the original problem statements to ensure they hold true. | Check if \(P_A + 2P_B\) equals 17 with calculated values. |
Additional Information: Linear Equations in Real Life
Linear equations and systems of linear equations are fundamental tools used to model and solve problems in many real-world situations, not just product pricing. Here are a few examples:
- Finance: Calculating interest, budgeting, analyzing costs and revenues.
- Physics: Describing motion, calculating forces, relating voltage, current, and resistance (Ohm's Law).
- Chemistry: Balancing chemical equations, calculating concentrations.
- Economics: Modeling supply and demand, analyzing costs.
- Mixture Problems: Determining the quantities of different substances needed to create a mixture with specific properties (like concentration or cost).
- Distance, Rate, Time Problems: Relating how far something travels based on its speed and duration.
Understanding how to set up and solve these equations is a crucial skill for solving a wide range of practical problems.
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