Solving the Electricity Bill Calculation Problem

This problem asks us to find the monthly electricity bill for a house, given that the bill depends on two factors: the number of rooms and the number of units consumed. We are provided with data from two different scenarios, which we can use to figure out the exact relationship between these factors and the bill.

According to the problem, the total monthly electricity bill consists of two parts:

• A part proportional to the number of rooms.
• A part proportional to the number of units consumed.

Let's represent the variables:

• \( B \) = Total monthly electricity bill
• \( R \) = Number of rooms
• \( U \) = Number of units consumed

Since the parts are proportional, we can write the relationship as a linear equation:

\( B = k_1 R + k_2 U \)

where \( k_1 \) is the constant of proportionality for the number of rooms and \( k_2 \) is the constant of proportionality for the number of units consumed.

We are given two scenarios:

1. House with 8 rooms and consuming 240 units has a bill of Rs. 400.
2. House with 6 rooms and consuming 200 units has a bill of Rs. 320.

We can use these scenarios to set up a system of two linear equations with two unknowns (\( k_1 \) and \( k_2 \)).

From scenario 1:

\( 400 = k_1 \times 8 + k_2 \times 240 \)

This simplifies to:

\( 8k_1 + 240k_2 = 400 \)

We can divide this equation by 8 to simplify it further:

\( k_1 + 30k_2 = 50 \) (Equation 1)

From scenario 2:

\( 320 = k_1 \times 6 + k_2 \times 200 \)

This simplifies to:

\( 6k_1 + 200k_2 = 320 \)

We can divide this equation by 2 to simplify it further:

\( 3k_1 + 100k_2 = 160 \) (Equation 2)

Now we need to solve this system of linear equations to find the values of \( k_1 \) and \( k_2 \).

From Equation 1, we can express \( k_1 \) in terms of \( k_2 \):

\( k_1 = 50 - 30k_2 \)

Substitute this expression for \( k_1 \) into Equation 2:

\( 3(50 - 30k_2) + 100k_2 = 160 \)

Distribute the 3:

\( 150 - 90k_2 + 100k_2 = 160 \)

Combine the \( k_2 \) terms:

\( 150 + (100 - 90)k_2 = 160 \)

\( 150 + 10k_2 = 160 \)

Subtract 150 from both sides:

\( 10k_2 = 160 - 150 \)

\( 10k_2 = 10 \)

Divide by 10 to find \( k_2 \):

\( k_2 = \frac{10}{10} \)

\( k_2 = 1 \)

Now substitute the value of \( k_2 \) back into the expression for \( k_1 \):

\( k_1 = 50 - 30k_2 \)

\( k_1 = 50 - 30(1) \)

\( k_1 = 50 - 30 \)

\( k_1 = 20 \)

So, the constants of proportionality are \( k_1 = 20 \) and \( k_2 = 1 \). This means the formula for the monthly electricity bill is:

\( B = 20R + 1U \)

or simply

\( B = 20R + U \)

Now we can use this formula to find the monthly electricity bill for a house with 7 rooms consuming 300 units.

Here, \( R = 7 \) and \( U = 300 \). Substitute these values into the formula:

\( B = 20(7) + 300 \)

\( B = 140 + 300 \)

\( B = 440 \)

The monthly electricity bill for a house with 7 rooms consuming 300 units is Rs. 440.

Let's check our constants with the given scenarios:

• Scenario 1: \( B = 20(8) + 240(1) = 160 + 240 = 400 \) (Matches the given Rs. 400)
• Scenario 2: \( B = 20(6) + 200(1) = 120 + 200 = 320 \) (Matches the given Rs. 320)

The formula is correct.

The calculation for 7 rooms and 300 units is:

\( B = 20(7) + 300 = 140 + 300 = 440 \)

Therefore, the monthly electricity bill is Rs. 440.

This type of problem demonstrates how real-world scenarios involving proportional relationships can be modeled and solved using systems of linear equations.

Revision Table: Electricity Bill Calculation

Additional Information: Understanding Proportionality and Costs

In many real-world cost calculations, expenses can be broken down into components that depend on different factors. These dependencies are often modeled using proportionality.

Direct Proportionality:

• When a quantity \( y \) is directly proportional to a quantity \( x \), it means that as \( x \) increases, \( y \) increases at a constant rate. The relationship is written as \( y = kx \), where \( k \) is the constant of proportionality.
• In our electricity bill problem, the part of the bill related to rooms is directly proportional to the number of rooms (\( k_1 R \)), and the part related to units is directly proportional to the number of units consumed (\( k_2 U \)).

Setting up equations from word problems:

• Identify the unknown quantities you need to find (in this case, the constants \( k_1 \) and \( k_2 \)).
• Use the information given in the problem to write equations that relate the known values (bill amounts, room numbers, unit consumption) to the unknown quantities.
• If there are two unknown quantities, you typically need two independent equations to solve for them. This leads to a system of equations.

Solving systems of equations is a fundamental skill in algebra and is applicable to many problems involving multiple variables and relationships.