Question 4

This problem involves understanding how a balance decreases at a constant rate over time.

Step 1 — Model the balance

Let's define the variables. Let $b(x)$ be the remaining balance. Let $x$ be the number of days. The initial recharge amount is ₹600. The balance reduces by ₹15 each day. So, the total reduction after $x$ days is $15x$. The remaining balance $b(x)$ is the initial amount minus the total reduction.

$b(x) = 600 - 15x$

This equation represents linear decay. The balance decreases by a constant amount. It decreases by ₹15 every single day. This constant rate of change shows linear decay.

Step 2 — Calculate days to run out

The balance runs out when it becomes zero. We set the remaining balance $b(x)$ to 0.

$b(x) = 0$

$600 - 15x = 0$

Let's solve this equation for $x$. We add $15x$ to both sides of the equation.

$600 = 15x$

Now, we divide both sides by 15.

$x = \frac{600}{15}$

$x = 40$

$\boxed{40 \text{ days}}$

The balance will run out after 40 days.

Step 3 — Create the table of values

We will calculate the balance $b(x)$ for $x$ from 1 to 10 days. We use the equation $b(x) = 600 - 15x$.

For $x = 1$: $b(1) = 600 - 15(1) = 585$

For $x = 2$: $b(2) = 600 - 15(2) = 570$

We continue this calculation for other values of $x$.

| x (days) | b(x) (₹) | | :------- | :------- | | 1 | 585 | | 2 | 570 | | 3 | 555 | | 4 | 540 | | 5 | 525 | | 6 | 510 | | 7 | 495 | | 8 | 480 | | 9 | 465 | | 10 | 450 |

The table shows how the balance reduces with time.

Answer

(i) The equation that models the remaining balance $b(x)$ after $x$ days is $b(x) = 600 - 15x$. This represents linear decay because the balance decreases by a constant amount (₹15) every day. (ii) The balance will run out after 40 days. (iii) | x (days) | b(x) (₹) | | :------- | :------- | | 1 | 585 | | 2 | 570 | | 3 | 555 | | 4 | 540 | | 5 | 525 | | 6 | 510 | | 7 | 495 | | 8 | 480 | | 9 | 465 | | 10 | 450 |