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The population of a village increase by 5% annually. If its present population is 4410, then its population 2 years ago was:
Answer & Solution
Correct Answer:
Option
B
Using compounding formula
$${\text{a}} + {\text{b}} + \frac{{{\text{ab}}}}{{100}}$$
we have
$$\eqalign{ & 5 + 5 + \frac{{25}}{{100}} \cr & = 10.25\% \cr} $$
so if the population 2 yrs ago be x
then
$$\eqalign{ & {\text{x}} + \frac{{10.25x}}{{100}} = 4410 \cr & {\text{or}},\,110.25{\text{x}} = 441000 \cr & \therefore {\text{x}} = 4000 \cr} $$
$${\text{a}} + {\text{b}} + \frac{{{\text{ab}}}}{{100}}$$
we have
$$\eqalign{ & 5 + 5 + \frac{{25}}{{100}} \cr & = 10.25\% \cr} $$
so if the population 2 yrs ago be x
then
$$\eqalign{ & {\text{x}} + \frac{{10.25x}}{{100}} = 4410 \cr & {\text{or}},\,110.25{\text{x}} = 441000 \cr & \therefore {\text{x}} = 4000 \cr} $$
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LoginLet it be 100,
For 1st yr is 105
2nd yr is 110.25
So, 4410-110.25 is approx 4300
population in the last year = 4410-220.5 = 4189.5
for the 2nd year = (5/100)*4189.5 = 209.475
population in the 2nd last year = 4189.5-209.475 = 3980.025
according to my calculation using the concept of increase.
again 2nd year 5% increase 4200+210=4410.
its like a compound interest.