Question

The population of a city is increased by 6%in 2004,by 4%in 2005, and 2%in 2006.if the population was 140556 in the beginning of 2007, what was the population as the start of 2004?

Answer 1

$\textbf{Given:}$

$\begin{array}{|c|c|}\cline{1-2}\textsf{Year}&\textsf{Rate of population increasing }\\\cline{1-2}2004&6\%\\2005&4\%\\2006&2\%\\\cline{1-2}\end{array}$

$\textbf{To find:}$

$\textsf{The population of the city in 2004}$

$\textbf{Solution:}$

$\underline{\textbf{Compound interest formula for different rates:}}$

$\boxed{\mathsf{A=P\left(1+\dfrac{a}{100}\right)\left(1+\dfrac{b}{100}\right)\left(1+\dfrac{c}{100}\right)}}$

$\textsf{Let P be the population in 2004}$

$\mathsf{Using\;compound\;interest\;formula}$

$\mathsf{A=P\left(1+\dfrac{a}{100}\right)\left(1+\dfrac{b}{100}\right)\left(1+\dfrac{c}{100}\right)}$

$\implies\mathsf{140556=P\left(1+\dfrac{6}{100}\right)\left(1+\dfrac{4}{100}\right)\left(1+\dfrac{2}{100}\right)}$

$\implies\mathsf{140556=P\left(\dfrac{106}{100}\right)\left(\dfrac{104}{100}\right)\left(\dfrac{102}{100}\right)}$

$\implies\mathsf{140556=P\left(\dfrac{53}{50}\right)\left(\dfrac{52}{50}\right)\left(\dfrac{51}{50}\right)}$

$\implies\mathsf{P=\dfrac{140556{\times}50{\times}50{\times}50}{53{\times}52{\times}51}}$

$\implies\mathsf{P=\dfrac{140556{\times}50{\times}50{\times}50}{140556}}$

$\implies\mathsf{P=50{\times}50{\times}50}$

$\implies\boxed{\mathsf{P=125000}}$

$\textbf{Answer:}$

$\textsf{Population of the city in 2004 is 125000}$

Answer 2

The population at the start of 2004 was 125496

Step-by-step explanation:

Let the population at the start of 2004 be = x

Thus population at the end of 2004 be = $x+\frac{6}{100}\times x=1.06\times x$

Then population at the end of 2005 be = $1.06\times x+\frac{4}{100}\times 1.06x=1.10\times x$

Then population at the end of 2006 be = $1.10\times x+\frac{2}{100}\times 1.10x=1.12\times x$

As it is given :

population in the beginning of 2007 = 140556  

Aspopulation at the end of 2006 = population in the beginning of 2007 = 140556

$1.12\times x=140556$

$x=125496$

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