S Chand ICSE Mathematics Class 9 Solution Second Chapter Compound Interest Exercise 2D (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

S Chand ICSE Mathematics Class 9 Solution Second Chapter Compound Interest Exercise 2D

(1) The cost of a machine depreciates by 10% every year. If its present worth is Rs. 18000, what will be its value after 3 years?


(2) The population of a town increases by 20% every year. If its present population is 2,16,000. Find its population (i) after 2 years (ii) 2 years ago.

(3) The machinery of a certain factory is valued at Rs. 18,400 at the end of 1980. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value of the machine at the end of 1979 and 1981.


Let the value of the machine at the end of 1980 = 18400 (1 – 8/100)

= 18400 x 92/100

= 16,928 Rs.

(4) The present value of a scooter is Rs. 15,360. If its value depreciates 12 and 1/2% every year, find its value after 3 years.

(5) a new car is purchased for Rs. 2,50,000. Its value depreciates at the rate of 10% in the first year, 8% in the 2nd year and then 6% every year. Find the value after 4 years.

(6) The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. If the original count of the bacteria in a sample is 10,000, find the bacteria count at the end of 3 hours.


(7) The production of refrigerators in a factory rose from 40000 to 48400 in 2 years. find the rate of growth p.a.

(8) The value of a flat worth Rs. 5,00,000 is depreciating at the rate of 10% p.a. In how many years will its value be reduced to Rs. 3,64,500?


(9) Rachit bought a flat for Rs. 10 lakh and a car for Rs. 3,20,000 at the same time. The price of the flat appreciates uniformly at the rate of 20% p.a., while the price of the car depreciates at the rate of 15% p.a. If Rachit sells the flat and car after 3 years, what will be his profit or loss?

Solution: Total C.P. of car and flat = (10,000,00 + 3,20,000) Rs.

= 13,20,000 Rs.

(10) 8,000 workers were employed by a company to complete a job in 4 years. At the end of 1st year, 5% of the worker were retrenched. At the end of 2nd year, 5% of those working at that time were retrenched. However to complete the job in time, the number of workers was increased by 10% of those working at the end of third year. How many workers were working during the fourth year?

Solution:  At the end of first year number of workers = 8000 x 19/20
= 7600
At the end of 2nd year number of workers = 7600 x 19/20
= 7220
At the end of 3rd year number of workers = 7220 x 11/10
= 7942

_____ x _____


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  1. Sir I need complete solution of class 9 maths..from where did I get it..

  2. Ans 7 is r_10% your is 20% which is wrong

  3. 22/20-1/1=r/100

  4. Solution of chapter test

  5. I want all soutions


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