Mathematics
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.
Polynomials
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Answer
Initial value of the mobile phone = ₹10,000
Decrease in value per year = ₹800
(i) Value of the phone after 3 years = 10000 - 800 × 3
= 10000 - 2400
= ₹7600
∴ The value of the phone after 3 years is ₹7600.
(ii) The value of the phone at the end of t years is given by v = 10000 - 800t.
| Year, t | Value, v (₹) |
|---|---|
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
(iii) The expression relating v and t is:
v = 10000 - 800t
This represents linear decay because as t (years) increases by 1, the value v decreases by a constant amount of ₹800. The change in v for every unit change in t is the same (a fixed decrease), which is the characteristic feature of linear decay.
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