Class 8 Maths Comparing Quantities Application of Compound Interest Formula

Application of Compound Interest Formula

There are some situations where we could use the formula for calculation of amount in CI.

 

(i) Increase (or decrease) in population.

 

(ii) The growth of a bacteria if the rate of growth is known.

 

(iii) The value of an item, if its price increases or decreases in the intermediate years.

 

Problem: The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(i) Find the population in 2001.                   (ii) What would be its population in 2005?

Solution:

(i) Here, A2003 = 54,000, R = 5%, n = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

So, A2003 = P2001(1 + R/100)n

 => 54000 = P2001(1 + 5/100)2

=> 54000 = P2001(1 + 1/20)2

=> 54000 = P2001(21/20)2

=> 54000 = P2001 * 21/20 * 21/20

=> P2001 = (54000 * 20 * 20)/(21 * 21)

=> P2001 = 48,980 (approx.)

(ii) According to question, population is increasing.

Therefore population in 2005,

A2005 = P(1 + R/100)n

         = 54000(1 + 5/100)2

         = 54000(1 + 1/20)2

         = 54000(21/20)2

         = 54000 * 21/20 * 21/20

         = 59,535

Hence population in 2005 would be 59,535.

Problem: In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Class_8_Maths_Comparing_Quantities__Example_5

Solution:

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P(1 + R/100)n

                    = 506000(1 + 2.5/100)2

                    = 506000(1 + 25/1000)2    

                    = 506000(1 + 1/40)2

                    = 506000(41/40)2

                    = 506000 * 41/40 * 41/40

                    = 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

 

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