Understanding the Cost Problem

This question asks us to find the price of a single bag based on two pieces of information relating the cost of bottles and bags. We are given that 5 bottles cost the same as 2 bags, and the total cost of 15 bottles and 4 bags is Rs. 2,000.

Setting Up Equations

To solve this, we can use algebra. Let's represent the cost of one bottle as \(B\) (in Rs.) and the cost of one bag as \(A\) (in Rs.). We can translate the given information into two equations:

1. The cost of 5 bottles is equal to the cost of 2 bags. This can be written as:
   \(5 \times \text{Cost of 1 bottle} = 2 \times \text{Cost of 1 bag}\)
   \(\text{Equation 1: } 5B = 2A\)
2. The total cost of 15 bottles and 4 bags is Rs. 2,000. This can be written as:
   \((15 \times \text{Cost of 1 bottle}) + (4 \times \text{Cost of 1 bag}) = 2000\)
   \(\text{Equation 2: } 15B + 4A = 2000\)

Now we have a system of two linear equations with two variables (\(B\) and \(A\)).

• Equation 1: \(5B = 2A\)
• Equation 2: \(15B + 4A = 2000\)

Solving the System of Equations

We need to find the value of \(A\) (the price of a single bag). We can use the substitution method.

1. From Equation 1, we can express \(B\) in terms of \(A\):
   Divide both sides of \(5B = 2A\) by 5:
   \(B = \frac{2A}{5}\)
2. Substitute this expression for \(B\) into Equation 2:
   Replace \(B\) with \(\frac{2A}{5}\) in the equation \(15B + 4A = 2000\):
   \(15 \left(\frac{2A}{5}\right) + 4A = 2000\)
3. Simplify the equation:
   \( \frac{15 \times 2A}{5} + 4A = 2000 \)
   \( \frac{30A}{5} + 4A = 2000 \)
   \( 6A + 4A = 2000 \)
4. Combine the terms with \(A\):
   \( 10A = 2000 \)
5. Solve for \(A\):
   Divide both sides by 10:
   \( A = \frac{2000}{10} \)
   \( A = 200 \)

So, the price of a single bag is Rs. 200.

Conclusion on Bag Price

Based on our calculations, the cost of a single bag is Rs. 200.

Revision Table: Cost Problem Analysis

Additional Information: Solving System of Equations

A system of linear equations involves two or more linear equations with the same set of variables. The solution to the system is the set of values for the variables that satisfy all equations simultaneously.

Common methods for solving systems of linear equations include:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This reduces the system to a single equation with one variable. This is the method we used here.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then add the equations together to eliminate that variable. The resulting equation has only one variable.
• Graphical Method: Graph each equation on the same coordinate plane. The point where the lines intersect represents the solution to the system. This method is often less precise than algebraic methods for non-integer solutions.

In this problem, the substitution method was straightforward because Equation 1 allowed us to easily isolate \(B\) in terms of \(A\).