Exercise 7.3
- 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?
Population in 2003 = Population in 2001 ×
Population in 2001 = 400 × 54000 441 = (round off to the nearest number)
The population in 2001 is 48980.
(ii) Population in 2005 = Population in 2003 × 1 + R 100 n = ×
= 54000 × 21 20 2 = 54000 × 441 400 = 135 × 441 =
The population in 2005 is 59535.
- 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.
Number of bacteria at the end of 2 hours = Number of count of bacteria initially ×
= × 1 + 2.5 100 2 = 506000 ×
= 506000 × = (Round off to the nearest number)
Thus, the number of bacteria after two hours = 5,31,616 (approx).
- A scooter was bought at Rs. 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Value of the scooter (after 1 yr) = ×
= × 1 − 8 100 1 = 42000 × = Rs.
Hence, the value of scooter after 1 year = ₹ 38,640.