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Problems on Population Growth

Comparing Quantities III - Concepts

Simple Interest

Compound Interest Compounded Annually

Compound Interest Different Rates

Compound Interest Time in Fractions

Compound Interest Half Yearly

Compound Interest Quarterly

Inverse Problems on Compound Interest - Principal

Inverse Problems on Compound Interest - Rate

Inverse Problems on Compound Interest Time Period

Problems on Population Growth

Problems on Depreciation

Class - RPF Constable Subjects

Concept Explanation

Problems on Population Growth

In our day to day life we observe that there are things or entities like population of city, the value of property, the height of a tree, weight and height of a child, the number of bacteria etc. which increase in magnitude over a period of time.

Growth: The relative increase in a quantity or entity is called growth .

Rate of Growth: Growth per unit of time is called the rate of growth.

Formulae for population Growth:

Formula 1: Let P be the population of a city or a town at the beginning of a certain year and the population grows at a constant rate of R% per annum, then

Population after n years = $Pleft ( 1+frac{R}{100} right )^{n}$

Formula 2: Let P be the population of a city or a town at the beginning of a certain year. if the population grows at the rate of $R_{1}%$ during second year, then

Population after 2 years = $Pleft ( 1+frac{R_{1}}{100} right )times left ( 1+frac{R_{2}}{100} right )$

This formula may be extended for more than 2 years.

Formula 3: Let P be the population of a city or a town at the beginning of a certain year. If the population decreases at the rate of R% per annum, then

Population after n years = $Pleft ( 1-frac{R}{100} right )^{n}$

Illustration: The population of a village is 20000. If the annual birth rate is 4 % and the annual death rate 2%., calculate the population after two years.

Solution: We have,

Annual birth rate = 4%

Annual death rate = 2%

So, Annual growth = ( 4 - 2) = 2 %

Thus, we have

P = Initial population = 20000, Rate of growth = 2% per annum, n = 2 years

So, Population after 2 years = $Pleft ( 1+frac{R}{100} right )^{n}$

$=20000left ( 1+frac{2}{100} right )^{2}$

$=20000left ( 1+frac{1}{50} right )^{2}$

$=20000left ( frac{51}{50} right )^{2}$

$=20000times frac{51}{50}times frac{51}{50}$

$=20808$

Hence population of the town after 2 years = 20808.

Sample Questions

(More Questions for each concept available in Login)

Question : 1

The population of a town was 160000 three years ago. If it has been increased by 3%, 2.5% and 5% in the last three years, find the present population of the town.
A. 177366	B. 17736	C. 17366	D. 173766
Right Option : A
View Explanation

Question : 2

The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and increases by 10% in the third hour.If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of three hours.
A. 10980	B. 10890	C. 1980	D. 18090
Right Option : B
View Explanation

Question : 3

A factory increased its production of three wheelers from 80000 in 1999 to 92610 in 2002. Find the annual rate of growth of production of three wheelers.
A. 15%	B. 25%	C. 10%	D. 5%
Right Option : D
View Explanation