Question 2

Let's find the phone's value after some years.

Step 1 — Calculate value after 3 years

The phone's initial value is ₹10,000. Its value decreases by ₹800 every year. First, we find the total decrease in value.

$\text{Total decrease} = \text{Decrease per year} \times \text{Number of years}$

$= 800 \times 3$

$= 2400$

The total decrease is ₹2,400. Now, we subtract this from the initial value.

$\text{Value after 3 years} = \text{Initial value} - \text{Total decrease}$

$= 10000 - 2400$

$\boxed{₹7600}$

Step 2 — Create a table of values

Let's make a table for $t$ from 0 to 8 years. We start with the initial value at $t=0$. Then, we subtract ₹800 for each passing year.

| t (years) | v (₹) | | :-------: | :---: | | 0 | 10000 | | 1 | 9200 | | 2 | 8400 | | 3 | 7600 | | 4 | 6800 | | 5 | 6000 | | 6 | 5200 | | 7 | 4400 | | 8 | 3600 |

Step 3 — Find the expression and explain linear decay

Let $v$ be the value of the phone. Let $t$ be the time in years. The initial value is ₹10,000. The value decreases by ₹800 each year. So, after $t$ years, the total decrease is $800t$. We subtract this from the initial value.

$v = 10000 - 800t$

This expression shows linear decay. The value decreases by a constant amount. It decreases by ₹800 every year. This constant rate of change defines a linear relationship.

Answer

(i) The value of the phone after 3 years is ₹7,600. (ii) | t (years) | v (₹) | | :-------: | :---: | | 0 | 10000 | | 1 | 9200 | | 2 | 8400 | | 3 | 7600 | | 4 | 6800 | | 5 | 6000 | | 6 | 5200 | | 7 | 4400 | | 8 | 3600 | (iii) The expression is $v = 10000 - 800t$. It represents linear decay because the value decreases by a constant amount (₹800) each year.