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?

Solution:

(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.

Solution:

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.

Solution:

(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?

Solution:

(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 _____


7 Comments

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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. Thakur Priyansh Singh

    22/20-1/1=r/100
    2/20*100=
    200/20=10pa

  4. Solution of chapter test

  5. I want all soutions

  6. HELLO SIR CAN YOU GIVE ME EXPANSIONS SOLUTION PLEASE URGENTLY

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