Class 9th Maths Ganita Manjari - Part I Chapter 2: Introduction to Linear Polynomials Exercise 4: Exercise Set 2.4 Question 2 (ii) Solution

Question

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

(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.

Accepted Answer

Recall the expression derived in the previous question for the mobile phone value $v$ and time $t$.
 	ext{$v = 10000 - 800t$} Find the phone value for each value of $t$ from 0 to 8 years using the linear expression.
 ${color:green}	extbf{Answer: The table shows that the phone value decreases by ₹800 every year.}{/color}$|||TABLE#START header=true headerBold=true headerAlign=center|center bodyAlign=center|center colWidths=50|50 headerBg=#e2e8f0 bodyBg=#ffffff|||COL#1 $t$ (years) COL#2 $v$ (₹)|||COL#1 0 COL#2 10000 - 800 × 0 = 10000|||COL#1 1 COL#2 10000 - 800 × 1 = 9200|||COL#1 2 COL#2 10000 - 800 × 2 = 8400|||COL#1 3 COL#2 10000 - 800 × 3 = 7600|||COL#1 4 COL#2 10000 - 800 × 4 = 6800|||COL#1 5 COL#2 10000 - 800 × 5 = 6000|||COL#1 6 COL#2 10000 - 800 × 6 = 5200|||COL#1 7 COL#2 10000 - 800 × 7 = 4400|||COL#1 8 COL#2 10000 - 800 × 8 = 3600|||TABLE#END

$t$ (years)	$v$ (₹)
0	10000 - 800 × 0 = 10000
1	10000 - 800 × 1 = 9200
2	10000 - 800 × 2 = 8400
3	10000 - 800 × 3 = 7600
4	10000 - 800 × 4 = 6800
5	10000 - 800 × 5 = 6000
6	10000 - 800 × 6 = 5200
7	10000 - 800 × 7 = 4400
8	10000 - 800 × 8 = 3600

Question: 2 (ii)

Answer