Question

0. According to 2011 Census, the
population of a rural town was found to
be 64000. The Census authority also
found that the population of this
particular town also had a growth of
5% per annum. In how many years
after 2011, did the population of this
town reach 740887​

Answer 1

Step-by-step explanation:

Given Question :-

According to 2011 Census, the population of a rural town was found to be 64000. The Census authority also found that the population of this particular town also had a growth of 5% per annum.

Correct Question :-

According to 2011 Census, the population of a rural town was found to be 64000. The Census authority also found that the population of this particular town also had a growth of 5% perannum. In how many years after 2011, did the population of this town reach 74088?

Solution :-

Given that :-

The total population of. rural town was found by Census 2011 = 64000

Growthing rate in the population = 5% per annum

Let the number of years be n

Total population after 2011 = 74088

We know that

The population is increasing per year by year

So we have to calculate by using Compound interest formula

We know that

A = P[1+(R/100)]^n

We have

P = 64000

A = 74088

R = 5%

On Substituting these values in the above formula then

=> 74088 = 64000×[1+(5/100)]^n

=> 74088 = 64000×[1+(1/20)]^n

=> 74088 = 64000[(20+1)/20]^n

=> 74088= 64000×(21/20)^n

=> 74088/64000 = (21/20)^n

=> (21/20)^n = 74088/64000

=> (21/20)^n =( 9261×8)/(8×8000)

=> (21/20)^n = 9261/8000

=>(21/20)^n = (21)³/(20)³

=> (21/20)^n = (21/20)³

On Comparing both sides then

=> n = 3

Therefore, n = 3 years

Answer:-

After 3 years from 2011 the population will be 74088.

Used formulae:-

• A = P[1+(R/100)]^n

Where,

• A = Amount
• P = Principle
• R = Rate of Interest
• n = Number of time the interest is calculated compounded.

• (a/b)^m = a^m/b^m