1.
Maths -Growth and Depreciation
• As all of you know, population in any country keeps on increasing continuously and cost of machines, buildings etc gradually
decrease with time from their initial costs. The former increase is Growth and the later decrease is Depreciation.
• Working rule to assess initial or final cost is almost similar to that of Compound Interest what you have done earlier.
• Let us consider initial value as Vi and final value as Vf . If annual rate of growth be R,
Final value (Vf) = Vi (1 + R/100)T where T is duration of time in years.
If annual rate of Depreciation be R,
Final value (Vf) = Vi (1 – R /100)T where T is duration of time in years.
We will take a practical problem to understand the same.
**Population of a town was 125840 in the year 2017. If it had increased at 2.4% every year, find its population in i) 2019 ii) 2016.
Solution: i) Initial value Vi = 125840, R = 2.4 % , T = 2 years, Vf = ?
Population in 2019 (Vf) = Vi (1 + R/100)T
= 125840 (1 + 2.4/100)2 = 125840 x 1024/1000 x 1024/1000 = 131953 ( rounded off to whole number )
ii) Final value (Vf) = Vi (1 + R/100)T
125840 = Vi (1 + 2.4/100)1 [** As 2016 is before 2017, we will take the later as final value for ease.]
125840 = Vi x 1024/1000 i.e. Vi x1024 = 125840 x 1000
i.e. Vi = 125840 x 1000 / 1024 = 122891 (rounded off to whole number)
Now we will take a problem on Depreciation:

2.
Maths -Growth and Depreciation
• **A car was bought in 2013. If its value came to ₹545460 in the year 2015 and it depreciated by 16 % every year, at what
price it was bought? Also find its value in 2016.
• Solution:
• Initial value Vi = ?, R = 16 % , T = 2 years, Vf = 545460
Final value (Vf) = Vi (1 – R /100)T
545460 = Vi (1 – 16/100)2 i.e. 545460 = Vi x 84/100 x 84/100
i.e. 545460 x 100 x 100 = Vi x 84 x 84 i.e. Vi = 545460x100/84 x 100/84 = ₹ 773044 ( rounded off to Re.)
Value in 2016: Final value (Vf) = Vi (1 – R /100)T
i.e. Vf = 545460 ( 1 – 16/100)1
= 545460 x 84/100 = ₹ 458186 (rounded off to nearest Re.)