Chapter 8.A, Problem 1QQ

To determine

To Determine the population at the end of second year.

Answer to Problem 1QQ

Solution:

The change from in one year is (110,000-100,000) = 10000.

So again in next year the population will increase by 10000

So population at the end of year 2 will be

Population at end of year 1 + constant change

110,000 + 10000 = 120,000

This makes option (b) as correct answer.

Explanation of Solution

Given Info:

A town’s population increases in one year from 100,000 to 110,000.Population is growing linearly, at steady constant rate.

First we need to understand what liner growth is. When the things increases by same constant amount at each step or at each time interval so it will be called as linear growth.

For example.

If on Monday you started with 10 friends on face book, on Tuesday you have 14 friends, then on Wednesday you have 18 friends, then on Thursday you have 22 friends. So this will be called as linear growth.

On the same way in our question also they used the word linear growth. That means population will change by same constant amount.

Let us see what that constant amount is.

The difference = 110,000 – 100,000 = 10000

Because the starting population is 100,000 and population at the end of year 1 is 110,000 so that makes the difference of 10000.

Now when again the population will increase by 10000 so it will become 110,000+10000 = 120,000.

Formula used:

We did not use any formula we just used the definition of linear growth. Linear growth means change by same constant amount at each time interval.

Calculation:

The constant difference = 110,000 – 100,000 = 10000

Population at end of year 1 + constant change

110,000 + 10000 = 120,000

Conclusion:

Whenever the growth is linear that means the change will be same constant amount at each time interval.