A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for varying from 0 to 8 years and show how the value of the phone, , depreciates with time. (iii) Find an expression that relates and , and explain why it represents linear decay.

Introduction to Linear Polynomials | Exercise 2.4

Question 2

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

(i) Find the value of the phone after 3 years.

(ii) Make a table of values for $t$ varying from 0 to 8 years and show how the value of the phone, $v$ , depreciates with time.

(iii) Find an expression that relates $v$ and $t$ , and explain why it represents linear decay.

Solution

Understand the Question

A mobile phone is purchased for an initial value of $\text{₹}10{,}000$ (at time $t = 0$ ).
The value decreases (depreciates) by a constant amount of $\text{₹}800$ each year.
To find the value after $t$ years, subtract the total depreciation ( $800t$ ) from the initial value: $v = 10000 - 800t$ .

(i) Find the value of the phone after 3 years.

Step 1 · Calculate Value After 3 Years

Given initial value $= \text{₹}10{,}000$ and yearly decrease $= \text{₹}800$ .

\begin{aligned} \text{Total decrease} &= 800 \times 3 \\[0.6em] &= 2400 \end{aligned}

\begin{aligned} \text{Value after 3 years} &= 10000 - 2400 \\[0.6em] &= 7600 \end{aligned}

Answer

(i) $\text{₹}7{,}600$

(ii) Make a table of values for $t$ varying from 0 to 8 years and show how the value of the phone, $v$ , depreciates with time.

Step 1 · Create Table of Values

Starting at $t = 0$ with $v = 10000$ and subtracting $800$ for each subsequent year:

\begin{array}{|c|c|} \hline t \text{ (years)} & v \text{ (₹)} \\ \hline 0 & 10000 \\ \hline 1 & 9200 \\ \hline 2 & 8400 \\ \hline 3 & 7600 \\ \hline 4 & 6800 \\ \hline 5 & 6000 \\ \hline 6 & 5200 \\ \hline 7 & 4400 \\ \hline 8 & 3600 \\ \hline \end{array}

Answer

(ii) $\begin{array}{|c|c|}\hline t \text{ (years)} & v \text{ (₹)} \\\hline 0 & 10000 \\\hline 1 & 9200 \\\hline 2 & 8400 \\\hline 3 & 7600 \\\hline 4 & 6800 \\\hline 5 & 6000 \\\hline 6 & 5200 \\\hline 7 & 4400 \\\hline 8 & 3600 \\\hline\end{array}$

(iii) Find an expression that relates $v$ and $t$ , and explain why it represents linear decay.

Step 1 · Formulate Expression and Explain Linear Decay

Let $v$ be the value of the phone in rupees and $t$ be the time in years.

Total decrease in $t$ years $= 800t$ .

$v = 10000 - 800t$

This expression represents linear decay because the value decreases by a fixed constant amount ( $\text{₹}800$ ) each year, yielding a constant rate of change and a polynomial of degree $1$ in $t$ with a negative slope.

Answer

(iii) $v = 10000 - 800t$ ; it represents linear decay because the value decreases at a constant rate of $\text{₹}800$ per year.

Common Mistakes

Initial Time Index Error: Setting $t = 1$ for the initial purchase price of ₹10,000 instead of $t = 0$ .
Linear vs Exponential Decay: Treating depreciation as a percentage reduction rather than a constant subtraction of ₹800 per year.
Sign Error: Writing $v = 10000 + 800t$ instead of subtracting the annual depreciation ( $v = 10000 - 800t$ ).

Question 2

More questions in Exercise 2.4