A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation that models the remaining balance after using the scheme for days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for varying from 1 to 10 days and show how the balance reduces with time.

Understand the Question Initial Value: The prepaid balance starts at ₹. Rate of Change: The balance reduces by a fixed amount of ₹ each day. Linear Decay: When a quantity decreases by a constant rate per unit of time, it is modeled by a linear polynomial of the form . The balance runs out when the remaining balance becomes zero (). Part (i) Write an equation that models the remaining balance after using the scheme for days. Explain why it represents linear decay. Step 1 · Model the Balance Equation Given: Initial balance Daily reduction rate Total reduction after days Therefore, the remaining balance after days is: This represents linear decay because the balance decreases by a constant amount (₹) each day, corresponding to a constant negative rate of change (slope). Answer > (i) . It represents linear decay because the balance decreases at a constant rate of ₹ per day. Part (ii) After how many days will the balance run out? Step 1 · Find Days when Balance is Zero The balance runs out when . Answer > (ii) 40 days Part (iii) Make a table of values for varying from 1 to 10 days and show how the balance reduces with time. Step 1 · Calculate Balance for Days 1 to 10 Using : For : For : Continuing for all values from to : Answer > (iii) Common Mistakes Incorrect Sign in the Equation: Writing instead of . Since the prepaid balance is decreasing, the slope must be negative. Linear vs. Exponential Decay: Confusing constant deduction (a fixed subtraction of ₹ per day, which is linear decay) with a percentage deduction (which is exponential decay).

Introduction to Linear Polynomials | Exercise 2.4

Question 4

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

(i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay.

(ii) After how many days will the balance run out?

(iii) Make a table of values for $x$ varying from 1 to 10 days and show how the balance $b(x)$ reduces with time.

Solution

Understand the Question

Initial Value: The prepaid balance starts at ₹ $600$ .
Rate of Change: The balance reduces by a fixed amount of ₹ $15$ each day.
Linear Decay: When a quantity decreases by a constant rate per unit of time, it is modeled by a linear polynomial of the form $b(x) = \text{Initial} - (\text{Rate} \times x)$ .
The balance runs out when the remaining balance becomes zero ( $b(x) = 0$ ).

(i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay.

Step 1 · Model the Balance Equation

Given:

Initial balance $= ₹600$
Daily reduction rate $= ₹15\text{ per day}$
Total reduction after $x$ days $= 15x$

Therefore, 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 (₹ $15$ ) each day, corresponding to a constant negative rate of change (slope).

Answer

(i) $b(x) = 600 - 15x$ . It represents linear decay because the balance decreases at a constant rate of ₹ $15$ per day.

(ii) After how many days will the balance run out?

Step 1 · Find Days when Balance is Zero

The balance runs out when $b(x) = 0$ .

\begin{aligned} b(x) &= 0 \\[0.6em] 600 - 15x &= 0 \\[0.6em] 15x &= 600 \\[0.6em] x &= \dfrac{600}{15} \\[0.6em] x &= 40 \end{aligned}

Answer

(ii) 40 days

(iii) Make a table of values for $x$ varying from 1 to 10 days and show how the balance $b(x)$ reduces with time.

Step 1 · Calculate Balance for Days 1 to 10

Using $b(x) = 600 - 15x$ :

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

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

Continuing for all values from $x = 1$ to $10$ :

\begin{array}{|c|c|} \hline x \text{ (days)} & b(x) \text{ (₹)} \\ \hline 1 & 585 \\ \hline 2 & 570 \\ \hline 3 & 555 \\ \hline 4 & 540 \\ \hline 5 & 525 \\ \hline 6 & 510 \\ \hline 7 & 495 \\ \hline 8 & 480 \\ \hline 9 & 465 \\ \hline 10 & 450 \\ \hline \end{array}

Answer

(iii) $\begin{array}{|c|c|}\hline x \text{ (days)} & b(x) \text{ (₹)} \\\hline 1 & 585 \\\hline 2 & 570 \\\hline 3 & 555 \\\hline 4 & 540 \\\hline 5 & 525 \\\hline 6 & 510 \\\hline 7 & 495 \\\hline 8 & 480 \\\hline 9 & 465 \\\hline 10 & 450 \\\hline\end{array}$

Common Mistakes

Incorrect Sign in the Equation: Writing $b(x) = 600 + 15x$ instead of $b(x) = 600 - 15x$ . Since the prepaid balance is decreasing, the slope must be negative.
Linear vs. Exponential Decay: Confusing constant deduction (a fixed subtraction of ₹ $15$ per day, which is linear decay) with a percentage deduction (which is exponential decay).

Question 4

More questions in Exercise 2.4