Question

The population of a town has been increasing at the rate of 10% every year. The present population is 121000. What was the population two years ago?
(a) 100000
(b) 110000
(c) 120000
(d) 122000

Answer 1

Rate of increase=10%
Present population=121000
So,population before 2 years,
$P(1+ \frac{r}{100})^{n}$
$121000 {(1 - \frac{1}{10} )}^{2} \\ = 121000 \times \frac{9}{10} \times \frac{9}{10} \\ = 1210 \times 81 \\ = 98010$

Answer 2

Given  : population of a town is 121000

it increases at the rate of 10% per annum,

To Find :  population two years ago

(a) 100000

(b) 110000

(c) 120000

(d) 122000

Solution:

Let say population 2 years ago = P

Rate of increase = 10 %

Hence current population after 2 years

=> 121000  = P ( 1 + 10/100)²

=> 121000 = P ( 1.1)²

=> 121000 =  1.21P

=> P  = 100000

population 2 years ago = 100000

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