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The present population of a town is 7,98,600. If it increased at the rate of 10% every year, what was its population 3 years ago? - Mathematics

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Question

The present population of a town is 7,98,600. If it increased at the rate of 10% every year, what was its population 3 years ago?

Sum
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Solution

Given, present population = 798600

Rate of growth, r = 10% per annum

Time, n = 3 years

Let the population three years ago be x.

Present population = Population after n years `(1 + r/100)^n`, where r is the rate of growth

∴ `x(1 + 10/100)^3 = 798600`

⇒ `x(11/10)^3 = 798600`

⇒ `x = (798600 xx 10 xx 10 xx 10)/(11 xx 11 xx 11)`

⇒ `x = 600000`

Hence, the population three years ago is 600000.

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Chapter 2: Compound Interest - EXERCISE 2C [Page 27]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 2 Compound Interest
EXERCISE 2C | Q 2. | Page 27
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