According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2004 did the population of this town reach 74,088 ?

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Solution

Let the no. of years=n
According to the question

$P (1 + \frac{R}{100})^{n} = A$

$64000 (1 + \frac{5}{100})^{n} = 74088$

$(1 + \frac{5}{100})^{n} = \frac{74088}{64000}$
$(\frac{(21}{20})^{n} = (\frac{21}{20})^{3}$

Equating the powers
n=3

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According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2004 did the population of this town reach 74,088 ?

Let the no. of years=n According to the question P(1+R100)n=A 64000(1+5100)n=74088 (1+5100)n=7408864000 ((2120)n=(2120)3 Equating the powers n=3

Let the no. of years=n
According to the question

$P (1 + \frac{R}{100})^{n} = A$

$64000 (1 + \frac{5}{100})^{n} = 74088$

$(1 + \frac{5}{100})^{n} = \frac{74088}{64000}$
$(\frac{(21}{20})^{n} = (\frac{21}{20})^{3}$

Equating the powers
n=3