Maths Class 8 Comparing Quantities Chapter 7 Exercise 7.3 Question Answer

Solutions For All Chapters Maths Class 8

Ex 7.3 Class 8 Maths Question 1.
The population of a place increased to 54000 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) It’s given that population in the year 2003 = 54,000

54,000 = (Population in 2001) (1 + 5/100)²

54,000 = (Population in 2001) (105/100)²

Population in 2001 = 54000 x (100/105)²

= 48979.59

Therefore, the population in the year 2001 was approximately 48,980

(ii) Population in 2005 = 54000(1 + 5/100)²

= 54000(105/100)²

= 54000(21/20)²

= 59535

Therefore, the population in the year 2005 would be 59,535.

Ex 7.3 Class 8 Maths Question 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.

Solution:

The initial count of bacteria is given as 5,06,000

Bacteria at the end of 2 hours = 506000(1 + 2.5/100)2

= 506000(1 + 1/40)2

= 506000(41/40)2

= 531616.25

Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).

Ex 7.3 Class 8 Maths Question32.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Principal = Cost price of the scooter = ₹ 42,000

Depreciation = 8% of ₹ 42,000 per year

= (P x R x T)/100

= (42000 x 8 x 1)/100

= ₹ 3360

Thus, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.