Page 132 - ICSE Math 8

Appreciation

In our daily life, many things and assets like value of property, population of city, etc., grow or appreciate over
a period of time. The relative increase in value is called appreciation and the appreciation per unit of time is
called the rate of appreciation.

(a) If P is the value of the asset at the beginning of a certain period and R% is the rate of appreciation/
Ê R ˆ n
growth per annum, then asset value after n year = P 1+ 100¯ ˜ .
Á
Ë
(b) If P is the value of the asset at the beginning of a certain year and R %, R %, ..., R % are the rates of
2
n
1
appreciation/growth in the 1st, 2nd, …, nth year respectively, then the asset value after n years
Ê R ˆ Ê R ˆ Ê R ˆ
n
1
2
= P 1+ 100¯ Ë 1+ 100¯ ... 1+ 100¯ ˜ .
˜
Á
Á
˜ Á
Ë
Ë
In some cases, the asset value reduces during certain years or rate of appreciation/growth is negative.
In such problems, the rate of appreciation/growth (R) is taken as negative, i.e., (–R) in the formula.
An example for such a case will be appreciation in population over the years. But the rate of appreciation
can turn negative over certain years due to, say, an epidemic.
Example 18: The population of a city was 4,60,000 in 2009. In 2010, it grew by 4%. In 2011, it dropped by
5%. In 2012, it again increased by 10%. What will be the population in 2013?
Solution: Let the population in 2013 be P.
Given, Initial population (P ) = 4,60,000, Growth rate in 2010 (R ) = 4%
1
1
Growth rate in 2011 (R ) = –5%, Growth rate in 2012 (R ) = 10%
2
3
Ê R ˆ Ê R ˆ Ê R ˆ
3
1
2
P = P 1+ 100¯ Ë 1+ 100¯ Ë 1+ 100¯ ˜
˜ Á
˜ Á
1Á
Ë
Ê 4 ˆ Ê ( - 5) ˆ Ê 10 ˆ Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ
fi P = 4,60,000 1+ 100¯ Ë 1+ 100 ¯ Ë 1+ 100¯ = 4,60,000 1+ 25¯ Ë 1- 20¯ Ë 1+ 10¯ ˜
˜
˜ Á
˜ Á
Á
˜ Á
˜ Á
Á
Ë
Ë
Ê 26ˆ Ê 19ˆ Ê 11ˆ
= 4,60,000 × Á 25¯ Ë 20¯ Ë 10¯ ˜ = 4,99,928
˜ Á
˜ Á
Ë
Hence, the population in 2013 will be 4,99,928.
Example 19: Given that carbon-14 (C ) decays at a constant rate in such a way that it reduces to 50% in
14
1
5,568 years. Find the age of an old wooden piece in which the carbon is only 12 % of the
original. 2
Solution: Let the rate of decay be R% per annum and the age of the wooden piece be n years.
Let the amount of carbon (originally) in the wooden piece be P.
P
Then, in 5,568 years, the amount left =
2
P Ê R ˆ 5568 1 Ê R ˆ 5568
\ = P 1- ˜ fi = 1- ˜ ...(1)
Á
Á
2 Ë 100¯ 2 Ë 100¯
1 25 P
After n years, the carbon left in the wooden piece = 12 % P = of P =
2 200 8
P Ê R ˆ n 1 Ê R ˆ n Ê 1ˆ 3 Ê R ˆ n
\ = P 1- ˜ fi = 1- ˜ fi Á ˜ = 1- ˜ ...(2)
Á
Á
Á
8 Ë 100¯ 8 Ë 100¯ Ë 2¯ Ë 100¯
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