the population of the town is increasing at tha rate of 6% p.a .it was 238765 in thr3 year 2018. find the population in the year 2016 and 2020.
Answers
If was the population is increasing at the rate of 6% and it becomes 238765 in the year 2018 then the population in the year 2016 was 212500 and in the year 2020 is 268276.35.
Step-by-step explanation:
The rate of population increase, R = 6% p.a.
The population in the year 2018 = 238765
We have to find the population for the years 2016 and 2020 which from 2018 will have a difference of 2 years.
∴ The time period for both the year 2016 & 2020, n = 2 years
Case (1): For the year 2016
Population for year 2018 = Population for the year 2016 * [1+ R/100]ⁿ
⇒ 238765 = Population for the year 2016 * [1+ 6/100]²
⇒ 238765 = Population for the year 2016 * [(106 * 106)/(100*100)]
⇒ Population for the year 2016 = [238765*100*100] / [106*106]
⇒ Population for the year 2016 = 212500
Case (2): For the year 2020
Population for year 2020 = Population for the year 2018 * [1+ R/100]ⁿ
⇒ Population for year 2020 = 238765* [1+ 6/100]²
⇒ Population for year 2020 = [238765*106*106] / [100*100]
⇒ Population for the year 2020 = 268276.35
Thus, the population in the year 2016 was 212500 and in the year 2020 is 268276.35.
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Let say Population in 2016 = X
Rate of increase per year = 6 %
For population in year 2018 , two years has been from 2016
Population in 2018 = X ( 1 + 6/100)²
= X (1.06)²
X (1.06)² = 238765
=> X = 212500
Population in 2016 = 212500
Population in 2020 = 212500 (1.06)⁴ = 268276
2016 212500
2017 225250
2018 238765
2019 253091
2020 268276