Question

5 प्रतिशत वार्षिक दर से बढ़ते हुए वर्ष 2005 के अंत में एक
हो गई। बताइए वर्ष 2003 में इस शहर की जनसंख्या कितना
एक जनित (COURT
चढ़त हुए वर्ष 2005 के अंत में एक शहर की जनसंख्या 44,100​

Answer 1

Answer:

The population in 2003 would be 40000.

Step-by-step explanation:

Let P be the population of the city in 2003,

Given,

Growth rate per year, r = 5% = 0.05,

Number of years from 2003 to 2005, t = 2,

Thus, the population at the end of 2005,

$A=P(1+r)^t$

$=P(1+0.05)^2$

$=P(1.05)^2$

According to the question,

$P(1.05)^2 = 44100$

$P=\frac{44100}{1.1025}=40000$

Hence, the population in 2003 would be 40000.

Learn more :

Find the rate of increasing of a population :

https://brainly.in/question/7780346

Population after decreasing  :

https://brainly.in/question/7199699

Answer 2

Given that,

Population at the end of 2005 = 44100

Growth rate per year r = 5%

Number of year 2003 to 2005 = 2

So, Time = 2 years

Let P be the population of the city in 2003,

We need to calculate the population at the end of 2005,

Using formula for population

$A=P(1+r)^{t}$

Where, t = time

P = population in 2003

A = population at the end of 2005

r = growth rate

Put the value into the formula

$44100=P(1+(0.05))^2$

$P=\dfrac{44100}{1.1025}$

$P=40000$

Hence, The population of the city in 2003 is 40000.