Question

The electricity bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed. When in a certain month 540 units are consumed, the bill is Rs. 1800. In another month 620 units are consumed and the bill is Rs.2040. In yet another month 500 units are consumed then, the bill for that month would be
A) Rs.1560
B) Rs.1680
C) Rs.1840
D) Rs.1950

Answer 1

Hint:
Here, we will assume the amount which is fixed and the amount which varies according to the number of units to be some variable. We will frame two linear equations with two variables and solve this further to find the amount fixed and the amount which is varying. Then we will use the obtained amounts to find the bill amount when 500 units are consumed.

Complete step by step solution:
We are given that the electricity bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed.
So, we know that the electricity bill is given by adding the amount which is fixed and the amount varying according to the number of units.
Let \[x\] be the electricity bill of a certain establishment which is partly fixed and$y$ be the electricity bill for one unit of electricity consumed which partly varies according to the number of units of electricity consumed.
We are given that in a certain month 540 units are consumed, and the corresponding electricity bill is Rs. 1800.
$x + 540y = 1800$ …………………………………………………………………………………..$\left( 1 \right)$
We are given that in another month 620 units are consumed, and the corresponding electricity bill is Rs. 2040.
$ \Rightarrow x + 620y = 2040$ …………………………………………………………………………………..$\left( 2 \right)$
Subtracting equation $\left( 1 \right)$ from equation $\left( 2 \right)$ , we get
$\left( {x + 620y} \right) - \left( {x + 540y} \right) = 2040 - 1800$
$ \Rightarrow 620y - 540y = 2040 - 1800$
Subtracting the like terms, we get
$ \Rightarrow 80y = 240$
Dividing by $80$ on both the sides, we get
$ \Rightarrow y = \dfrac{{240}}{{80}}$
$ \Rightarrow y = 3$
So, we get cost of one unit of electricity $y = {\text{Rs}}.3$
By substituting $y = {\text{Rs}}.3$ in equation $\left( 1 \right)$, we get
$x + 540\left( 3 \right) = 1800$
Multiplying the terms, we get
$ \Rightarrow x + 1620 = 1800$
Subtracting 1620 from both the sides, we get
$ \Rightarrow x = 1800 - 1620$
$ \Rightarrow x = 180$
So, we get the electricity bill is fixed $x = {\text{Rs}}.180$
Now, we will find the electricity bill for another month when 500 units are consumed by multiplying the number of units with the cost per unit and then adding with the electricity bill which is fixed.
Total Electricity Bill when 500 units are consumed $ = x + 500y$
Substituting $x = 180$ and $y = 3$ in the above equation, we get
$ \Rightarrow $ Total Electricity Bill when 500 units are consumed $ = 180 + 500\left( 3 \right)$
Multiplying the terms, we get
$ \Rightarrow $ Total Electricity Bill when 500 units are consumed $ = 180 + 1500$
Adding the terms, we get
$ \Rightarrow $ Total Electricity Bill when 500 units are consumed $ = {\text{Rs}}.1680$

Therefore, the bill amount when 500 units is consumed is ${\text{Rs}}.1680$. Thus Option (B) is the correct answer.

Note:
We know that the equation formed is a linear equation in two variables. Linear equation is an equation with the highest degree as 1 with two variables. Linear equations in two variables can be solved by using the method of substitution and the method of elimination. Here, we might make a mistake by multiplying 500 to the fixed amount which will give us the wrong answer. We must add the fixed amount to the variable amount or else we get the wrong answer.