Page 128 - Maths Skills - 8
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126 Maths
Let’s Attempt
Example 1: The population of a town is 35000 in a certain year. It increases at the rate of 4% per annum. What
will be its population after 2 years? Also find its net growth.
Solution: Let ‘P’ be the population at the beginning of a certain year. Then, the population after 2 years (A)
is given by
R n 4 2
A = P 1 = 35000 1
100 100
1 2 26 2 26 6
= 35000 1 = 35000 = 35000 × = 37856
25 25 25 25
Hence, the population of the town after 2 years will be 37856.
Net growth = A – P = 37856 – 35000 = 2856
Hence, the net growth in population will be 2856.
Example 2: The population of a town is 10000 in a certain year. It increases at the rate of 5% in the first year
and 10% in the second year. Find the net growth in two years.
Solution: Let the population at the beginning of a certain year be P = 10000.
R = 5% for the first year R = 10% for the second year.
1 2
Total population after 2 years
R R 5 10 105 110
A = P 1+ 1 1+ 2 = 10000 1+ 1+ = 10000 × × = 11550
100 100 100 100 100 100
Net growth = 11550 – 10000 = 1550
Hence, the net growth in two years is 1550.
Example 3: The cost of a production plant is ` 90000. If its value depreciates at 10% annually, find the value
of the plant after 4 years.
Solution: Let ‘P’ be the initial value, i.e., P = ` 90000
Depreciation R = 10% n = 4 years
Final value (A) of the plant after 4 years is given by
R n 10 4 9 4
A = P 1 = ` 90000 1 = ` 90000 ×
100 100 10
9 9 9 9
= ` 90000 × × × × = ` 59049
10 10 10 10
Hence, the value of the production plant after 4 years will be ` 59049.