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The present population of a town is 48400. It is increasing 10% every year. Find the population of the town: i. 2 years ago, ii. After 2 years. - Mathematics

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Question

The present population of a town is 48400. It is increasing 10% every year. Find the population of the town:

  1. 2 years ago, 
  2. After 2 years.
Sum
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Solution

Given:

  • Present population of the town (P0 = 48400)
  • Annual increase rate (r = 10%)

i. Population 2 years ago

Step 1: Let the population 2 years ago be P.

Step 2: Population grows by 10% each year, hence the present population after 2 years can be expressed as:

`P_0 = P xx (1 + r/100)^2`

P0 = P × (1.1)2

P0 = P × 1.21

Step 3: Solve for P:

`P = P_0/1.21`

`P = 48400/1.21`

P = 40000

ii. Population after 2 years

Step 1: Let the population after 2 years be P2.

Step 2: Population increases by 10% each year, so:

`P_2 = P_0 xx (1 + r/100)^2`

P2 = 48400 × (1.1)2

P2 = 48400 × 1.21

P2 = 58564

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Chapter 2: Compound Interest - Exercise 2C [Page 53]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 2 Compound Interest
Exercise 2C | Q 9. | Page 53
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