Question 2

We will form two linear equations and solve them.

Step 1 — Formulate equations

The monthly bill $y$ depends on the hours $x$. The relation is given as $y = ax + b$. When $x = \textbf{10}$ hours, the bill $y = \textbf{800}$ rupees. Let's substitute these values into the relation.

$800 = a(10) + b$

$10a + b = 800 \quad \text{(Equation 1)}$

When $x = \textbf{15}$ hours, the bill $y = \textbf{1100}$ rupees. Let's substitute these values into the relation.

$1100 = a(15) + b$

$\boxed{15a + b = 1100 \quad \text{(Equation 2)}}$

Step 2 — Solve for 'a'

We have two linear equations. Let's subtract Equation 1 from Equation 2. This will eliminate the variable $b$.

$(15a + b) - (10a + b) = 1100 - 800$

$15a - 10a + b - b = 300$

$5a = 300$

$a = \frac{300}{5}$

$\boxed{a = 60}$

Step 3 — Solve for 'b'

Now we know the value of $a$. Let's substitute $a = \textbf{60}$ into Equation 1. Equation 1 is $10a + b = 800$.

$10(60) + b = 800$

$600 + b = 800$

$b = 800 - 600$

$\boxed{b = 200}$

Answer

(i) The value of $a$ is $\textbf{60}$. (ii) The value of $b$ is $\textbf{200}$.