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Chapter 7: Comparing Quantities

    Introduction

    Finding Discounts

    Sales Tax/Value Added Tax/Goods and Services Tax

    Compound Interest

    Deducing a Formula for Compound Interest

    Applications of Compound Interest Formula

    Exercise 7.1

    Exercise 7.2

    Exercise 7.3

    What Have We Discussed?

    Enhanced Curriculum Support

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Chapter 7: Comparing Quantities > Exercise 7.3

Exercise 7.3

1. 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?

Instructions

Population in 2003 = Population in 2001 ×
= Population in 2001 × 1+51002 ⇒ 54000 = Population in 2001 × ⇒ 54000 = Population in 2001 ×
Population in 2001 = 400×54000441 = (round off to the nearest number)
The population in 2001 is 48980.
(ii) Population in 2005 = Population in 2003 × 1+R100n = ×
= 54000 × 21202 = 54000 × 441400 = 135 × 441 =
The population in 2005 is 59535.

2. 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.

Instructions

Number of bacteria at the end of 2 hours = Number of count of bacteria initially ×
= × 1+2.51002 = 506000 ×
= 506000 × = (Round off to the nearest number)
Thus, the number of bacteria after two hours = 5,31,616 (approx).

3. A scooter was bought at Rs. 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Instructions

Value of the scooter (after 1 yr) = ×
= × 1−81001 = 42000 × = Rs.
Hence, the value of scooter after 1 year = ₹ 38,640.