What is the monthly electricity bill for a house with m rooms and consuming n units?

Question

Prepp Team · Accepted Answer

Understanding the Electricity Bill Calculation

The problem states that the total monthly electricity bill for a house is made up of two parts. One part changes directly with the number of rooms in the house, and the other part changes directly with the number of units of electricity consumed.

Let's define the variables:

$B$ = Total monthly electricity bill
$m$ = Number of rooms
$n$ = Number of units consumed

Based on the problem description, we can write the general formula for the electricity bill as:

$B = k_1 m + k_2 n$

Here, $k_1$ is the constant of proportionality for the part of the bill related to the number of rooms, and $k_2$ is the constant of proportionality for the part of the bill related to the number of units consumed.

Setting Up Equations from Given Data

We are given two scenarios which provide specific values for the number of rooms, units consumed, and the corresponding total bill. We can use these to create two linear equations involving $k_1$ and $k_2$.

Scenario 1:

Number of rooms ($m$) = 8
Number of units consumed ($n$) = 240
Total bill ($B$) = Rs. 400

Substituting these values into our general formula:

$400 = k_1 \times 8 + k_2 \times 240$

This gives us our first equation:

$8k_1 + 240k_2 = 400$ (Equation 1)

We can simplify Equation 1 by dividing all terms by 8:

$k_1 + 30k_2 = 50$ (Simplified Equation 1)

Scenario 2:

Number of rooms ($m$) = 6
Number of units consumed ($n$) = 200
Total bill ($B$) = Rs. 320

Substituting these values into our general formula:

$320 = k_1 \times 6 + k_2 \times 200$

This gives us our second equation:

$6k_1 + 200k_2 = 320$ (Equation 2)

We can simplify Equation 2 by dividing all terms by 2:

$3k_1 + 100k_2 = 160$ (Simplified Equation 2)

Solving for the Constants k₁ and k₂

Now we have a system of two linear equations with two variables ($k_1$ and $k_2$):

$k_1 + 30k_2 = 50$
$3k_1 + 100k_2 = 160$

We can solve this system using methods like substitution or elimination.

Using the substitution method, let's solve Simplified Equation 1 for $k_1$:

$k_1 = 50 - 30k_2$

Now substitute this expression for $k_1$ into Simplified Equation 2:

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

Distribute the 3:

$150 - 90k_2 + 100k_2 = 160$

Combine the $k_2$ terms:

$150 + 10k_2 = 160$

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 $k_2 = 1$ back into the equation for $k_1$:

$k_1 = 50 - 30k_2$

$k_1 = 50 - 30(1)$

$k_1 = 50 - 30$

$k_1 = 20$

So, we have found the constants of proportionality: $k_1 = 20$ and $k_2 = 1$.

General Formula for Electricity Bill

Now that we have the values of $k_1$ and $k_2$, we can write the general formula for the monthly electricity bill for a house with $m$ rooms and consuming $n$ units:

$B = k_1 m + k_2 n$

Substitute $k_1 = 20$ and $k_2 = 1$:

$B = 20m + 1n$

$B = 20m + n$

The monthly electricity bill for a house with $m$ rooms and consuming $n$ units is Rs. $(20m + n)$.

Comparing with Options

Let's compare our derived formula with the given options:

Option 1: Rs. $(40 m + n)$
Option 2: Rs. $(20 m + n)$
Option 3: Rs. $\frac{(40 m+n)}{2}$
Option 4: Rs. $\frac{(30 m+n)}{2}$

Our formula, Rs. $(20m + n)$, matches Option 2.

Variable	Description	Derived Value
$m$	Number of rooms	Given
$n$	Number of units consumed	Given
$B$	Total monthly electricity bill	$20m + n$
$k_1$	Constant for room part	20
$k_2$	Constant for unit part	1

Electricity Bill Revision Table

Step	Action	Result
1	Define variables and general formula	$B = k_1 m + k_2 n$
2	Formulate Equation 1 from Scenario 1 (8 rooms, 240 units, Rs. 400)	$8k_1 + 240k_2 = 400$ (Simplified: $k_1 + 30k_2 = 50$)
3	Formulate Equation 2 from Scenario 2 (6 rooms, 200 units, Rs. 320)	$6k_1 + 200k_2 = 320$ (Simplified: $3k_1 + 100k_2 = 160$)
4	Solve the system of equations for $k_1$ and $k_2$	$k_1 = 20$, $k_2 = 1$
5	Substitute $k_1$ and $k_2$ into the general formula	$B = 20m + 1n$
6	Final formula for bill with $m$ rooms and $n$ units	Rs. $(20m + n)$

Additional Information on Proportionality and Linear Equations

Proportional Relationships:

A quantity $A$ is proportional to another quantity $B$ if $A = k \times B$, where $k$ is a constant of proportionality. In this problem, the electricity bill has two parts, each proportional to a different factor (rooms and units). This leads to a linear combination of the factors.

Systems of Linear Equations:

When we have two or more linear equations involving the same set of variables, it's called a system of linear equations. To find the values of the variables, we need to solve the system. Common methods include:

Substitution Method: Solve one equation for one variable, and then substitute that expression into the other equation.
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.

In this problem, we used the substitution method to find the constants $k_1$ and $k_2$.

Prepp Team · Answer

Rs. (40 m + n)

Prepp Team · Answer

Rs. $\rm \frac{(40 m+n)}{2}$

Prepp Team · Answer

Rs. $\rm \frac{(30 m+n)}{2}$

Step	Action	Result
1	Define variables and general formula	\(B = k_1 m + k_2 n\)
2	Formulate Equation 1 from Scenario 1 (8 rooms, 240 units, Rs. 400)	\(8k_1 + 240k_2 = 400\) (Simplified: \(k_1 + 30k_2 = 50\))
3	Formulate Equation 2 from Scenario 2 (6 rooms, 200 units, Rs. 320)	\(6k_1 + 200k_2 = 320\) (Simplified: \(3k_1 + 100k_2 = 160\))
4	Solve the system of equations for \(k_1\) and \(k_2\)	\(k_1 = 20\), \(k_2 = 1\)
5	Substitute \(k_1\) and \(k_2\) into the general formula	\(B = 20m + 1n\)
6	Final formula for bill with \(m\) rooms and \(n\) units	Rs. \((20m + n)\)

College	Students	Mechanical	Civil	Computer Science	Applied
IIT Delhi	430	-	20%	-	10%
IIT Kanpur	350	20%	-	25%	-
IIT Bombay	-	20%	18%	-	32%
IIT Madras	-	-	25%	18%	35%
IIT Guwahati	400	20%	22%	-	20%