The population of a city increases at compounding rate of 8% per year. Find the population in the year 2012 if population in the year 2010 was 2,50,000.
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The population of a city increases at compounding rate of 8 %
e.g., interest rate , r = 8 %
period , n = 2012 - 2010 = 2 yrs
population in the year 2010 was 2,50,000
e.g., P = 2,50,000
Use formula,![\bf{A= P\left[\begin{array}{c}1+\frac{r}{100}\end{array}\right]^n}](https://tex.z-dn.net/?f=%5Cbf%7BA%3D+P%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%2B%5Cfrac%7Br%7D%7B100%7D%5Cend%7Barray%7D%5Cright%5D%5En%7D "\bf{A= P\left[\begin{array}{c}1+\frac{r}{100}\end{array}\right]^n}")
A = 2,50,000 × (1 + 8/100)²
= 2,50,000 × (1.08)²
= 2,50,000 × 1.08 × 1.08
= 2,91,600
hence, population in the year 2012 = 2,91,600
e.g., interest rate , r = 8 %
period , n = 2012 - 2010 = 2 yrs
population in the year 2010 was 2,50,000
e.g., P = 2,50,000
Use formula,
A = 2,50,000 × (1 + 8/100)²
= 2,50,000 × (1.08)²
= 2,50,000 × 1.08 × 1.08
= 2,91,600
hence, population in the year 2012 = 2,91,600
Answered by
10
Answer:
The population of a city increases at compounding rate of 8 %
e.g., interest rate , r = 8 %
period , n = 2012 - 2010 = 2 yrs
population in the year 2010 was 2,50,000
e.g., P = 2,50,000
Use formula, \bf{A= P[\begin{array}{c}1+\frac{r}{100}\end{array}]^n}A=P[1+100r]n
A = 2,50,000 × (1 + 8/100)²
= 2,50,000 × (1.08)²
= 2,50,000 × 1.08 × 1.08
= 2,91,600
hence, population in the year 2012 = 2,91,600
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