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, A_{2003} = 54,000, R = 5%, n = 2 years

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

Here population is increasing.

So, A_{2003} = P_{2001}(1 + R/100)^{n}

=> 54000 = P_{2001}(1 + 5/100)^{2}

=> 54000 = P_{2001}(1 + 1/20)^{2}

=> 54000 = P_{2001}(21/20)^{2}

=> 54000 = P_{2001} * 21/20 * 21/20

=> P_{2001} = (54000 * 20 * 20)/(21 * 21)

=> P_{2001} = 48,980 (approx.)

(ii) According to question, population is increasing.

Therefore population in 2005,

A_{2005} = 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.

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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