Question

The present population of a city is 9261000. If it has been increasing at the rate of 5% per annum,
find its population 3 years ago.​

Answer 1

Answer :-

Population of city 3 years ago was 8,000,000

$\rule{50}2$

Given:-

• Present population of city (Anount = A) = 9261000
• Increased Rate (R) = 5%
• Time (t) = 3 years

Find:-

Population of city 3 years ago

Solution:-

We know that

$\sf{Amount\:=\:Principe\bigg(1\:+\:\dfrac{Rate}{100}\bigg)^{Time}}$

Or

$\sf{A\:=\:P\bigg(1\:+\:\dfrac{R}{100}\bigg)^{t}}$

Substitute the known values in above formula

=> $\sf{9261000\:=\:P\bigg(1\:+\:\dfrac{5}{100}\bigg)^{3}}$

=> $\sf{9261000\:=\:P\bigg(\dfrac{100\:+\:5}{100}\bigg)^{3}}$

=> $\sf{9261000\:=\:P\bigg(\dfrac{105}{100}\bigg)^{3}}$

=> $\sf{9261000\:=\:P\bigg(\dfrac{21}{20}\bigg)^{3}}$

=> $\sf{9261000\:=\:P\bigg(\dfrac{21}{20}\:\times\:\dfrac{21}{20}\:\times\:\dfrac{21}{20}\bigg)}$

=> $\sf{9261000\:=\:P\bigg(\dfrac{9261}{8000}\bigg)}$

=> $\sf{\dfrac{9261000}{9261}\:=\:P\bigg(\dfrac{1}{8000}\bigg)}$

=> $\sf{1000(8000)\:=\:P}$

=> $\sf{P\:=\:8000000}$

Answer 2

Question :- The present population of a city is 9261000. If it has been increasing at the rate of 5% per annum,find its population 3 years ago.

Answer :-

$given \: \\ the \: present \: population \: \: of \: the \: city \: is \: = 9261000 \\ increased \: rate = 5\% \\ time = 3 \: years \\ find \: the \: population \: 3 \: years \: ago =$