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Example 29: Find the time period if `8000 borrowed at compound interest at 20% per annum
amount to `8820, the interest being compounded quarterly.
20
Solution: A = `8820, P = `8000, R = 20% p.a. = % quarterly = 5% quarterly.
4
Let time (T) = x.


5 
R 
Now, Amount (A) = P 1 + 100  4n ⇒ `8820 = `8000 1 + 100  4x ⇒ 8820 =    105  4x




100
8000
⇒ 441 =   21  4x ⇒   21  2 =   21  4x ⇒ 2 = 4x
400  20  20  20
2 1
⇒ x = = year = 6 months
4 2
So, the required time period is 6 months.
Application of Compound Interest Formula
• We can solve problems related to appreciation, i.e.,

(i) increase in population
(ii) the growth of bacteria, if the rate of growth is known
(iii) the value of an item, if its price increases in the intermediate years, etc., by using the formula

R 

A = P 1 + 100  n ,


where A is the appreciated value (or increased value), P is the original value, R% is the rate of
growth per year and n is the number of years.
• We can also solve problems related to depreciation, i.e.,
(i) decrease in population

(ii) the value of an item, if its price decreases
in the intermediate years, etc. If the Note: 1. Depreciation refers to a negative change
rate of depreciation is R% per year in the value, decreases with time and the
decrease of the value per year or unit of
and the initial value of item is P, the time is called rate of depreciation.
depreciated value at the end of n years 2. If there is depreciation in the value, then
 R  n rate of depreciation (R) is taken as negative.
= P1 − 100  .


• Many a times, the rate of growth becomes different in different times. In that case, we use a
different formula. If R , R , R , ..., R , are the growth rates respectively for the first, second,
n
1
2
3
third, ..., nth years, then the formula for the population of the town after n years (say A), whose
initial population is P given by
 R   R   R   R 
A = P 1 + 100  1 + 100  1 + 100 ...   1 + 100
2
3
1
n


 
 


Example 30: The population of a town is 1,20,000. If the annual birth rate is 6% and annual death
rate is 4%. Find the population after two years.
Solution: Annual birth rate = 6%, annual death rate = 4%
Annual growth = (6 – 4)% = 2%
Initial population (P) = 1,20,000
Mathematics-8 184