Questions with Answers, Solution | Life Mathematics | Chapter 4 | 8th Maths - Applications of Compound Interest Formula | 8th Maths : Chapter 4 : Life Mathematics
Chapter: 8th Maths : Chapter 4 : Life Mathematics
Applications of Compound Interest Formula
Applications
of Compound Interest Formula
The compound
interest formula is used in the following situations.
(i) To find
the increase 
or decrease 
in population.
(ii) To find
the growth of cells when the rate of growth is given.
(iii) To
find the depreciation in the values of machines, vehicles, utility appliances etc.,
Example 4.15
The value
of a motor cycle 2 years ago was ₹70000. It depreciates at the rate of 4% p.a.
Find its
present value.
Solution:
= ₹645121
Example 4.16
The bacteria
in a culture grows by 5% in the first hour, decreases by 8% in the second hour and
again increases by 10% in the third hour. Find the count of the bacteria at the
end of 3 hours, if its initial count was 10000.
Solution:
Bacteria
at the end of 3 hours
A = ₹10626
Example 4.17
The population
of a town is increasing at the rate of 6% p.a. It was 238765 in the year 2018. Find
the population in the year 2016 and 2020.
Solution:
Let the population in 2016 be P.
= 238765
× 53/50 × 53/50
= 95.506
× 53 × 53
A = 268276
The population
in the year 2016 was 212500 and that in the year 2020 will be 268276.
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