Question

the population of a place increased to 54000 in 2003 at a rate of 5% per annum a find the population in 2001 B what will be its population in 2005 ​

Answer 1

Given:

Population in 2003 =54000....(1).

Increase in population every year =5%.

Now, let the population in 2001 is x . Before we proceed, we should know that, what do we mean by a 5% increase.

Let us consider initially we have 100 numbers of pens. Then, 1% of 100 =1100×100=1.

So, 5% of 100 =5100×100=5.

If there is a 5% increase in the number of pens then, final number of pens = (Initial number of pens) + (5% of the initial number of pens).

⇒ Final number of pens =100+5=105.

Let us consider a number N . Then, y% of N = y100×N . If N increases by y% . Then, N will become N+y/100×N=N×(1+y/100).

In our question y=5 . Then, N will become N+y/100×N=N×(1+5/100)=N×1.05.

From the above calculation, if any number is increased 5% then, it’s final value will be 1.05 times the initial value. We will use this result directly to solve this question.

We have,

Population in 2001 =x.

Then, after a 5% increment population in 2002 =1.05x.

Similarly, after a 5% increment, the population in 2003 =1.05×(1.05x)=(1.05)2x(2).

Now, equating (1) and (2). Then,

(1.05)2x=54000⇒x=54000/(1.05)2=54000/1.1025=48979.59184          ⇒x≈48980

Thus, the population in 2001 is 48980 approximately.

Now, as a population in 2003 = 54000 . Then, after a 5% increment, the population in 2004 =1.05×54000=56700.

Similarly, after a 5% increment, the population in 2005 =1.05×56700=59535 .

Thus, the population in 2005 is 59535.

Now, the impact of high population is given below:

1. Problem of unemployment.

2. Adverse impact on the environment. For example: if the population is huge then more migration of a large number of people to urban areas with industrialization. This results in air, water and noise pollution in big cities and towns.

3. Per capita income will be reduced.