RD Sharma Solutions Class 8 Compound Interest Exercise 14.5

RD Sharma Class 8 Solutions Chapter 14 Ex 14.5 PDF Free Download

RD Sharma Solutions Class 8 Chapter 14 Exercise 14.5

Exercise 14.5

 Q1) Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5% per year, calculate its value after 2 years.

Solution:

Value of the boat after two years = \(P(1-\frac{R}{100})^{n}\)

=> \(16000(1-\frac{5}{100})^{2}\)

= 16000 (0.95)2

= 14,440

Thus, the value of the boat after two years will be Rs 14,440.

 

Q2) The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000? Also, find the total depreciation during this period.

Solution:

 Value of the machine after two years = \(P(1-\frac{R}{100})^{n}\)

=> \(100000(1-\frac{10}{100})^{2}\)

= 100000 (0.90)2

= 81,000

Thus, the value of the machine after two years will be Rs 81,000.

Depreciation = Rs 100,000 – Rs 81,000

= Rs 19,000

Q3) Pritam bought a plot of land for Rs 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years?

Solution:

Given:

P = Rs 64,000

R = 5% for every six months

Value of the plot after two years = \(P(1+\frac{R}{100})^{n}\)

=> \(64000(1+\frac{5}{100})^{4}\)

= \(64000(1.025)^{4}\)

= 706,440.25

Thus, the value of the plot after two years will be Rs 706,440.25.

Q4) Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.

Solution:

Value of the house after three years = \(P(1-\frac{R}{100})^{n}\)

=> \(30000(1-\frac{25}{100})^{3}\)

= \(30000(0.75)^{3}\)

= 12,656.25

Thus, the value of the house after three years will be Rs 12,656.25.

  

Q5) The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.

Solution:

Purchase price = \(P(1-\frac{R}{100})^{-n}\)

=> \(43740(1-\frac{10}{100})^{-3}\)

= \(43740(0.90)^{-3}\)

= 60,000

Thus, the purchase price of the machine was Rs 60,000.

 

Q6) The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?

Solution:

Purchase price = \(P(1-\frac{R}{100})^{-n}\)

=> \(9680(1-\frac{12}{100})^{-2}\)

= \(9680(0.88)^{-2}\)

= 12,500

Thus, the purchase price of the refrigerator was Rs 12,500.

 

Q7) The cost of a T.V. set was quoted Rs 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What was the cost of the T.V. set in 2001?

Solution:

Cost of the TV = \(P(1+\frac{R}{100})(1-\frac{R}{100})\)

=> \(17000(1+\frac{5}{100})(1-\frac{4}{100})\)

= 17,000 (1.05) (0.96)

= 17,136

Thus, the cost of the TV in 2001 was Rs 17,136.

Q8) Ashish started the business with an initial investment of Rs 500000. In the first year he incurred a loss of 4%. However during the second year he earned a profit of 5% which in third year rose to 10%. Calculate the net profit for the entire period of 3 years.

Solution:

Profit for three years = \(P(1-\frac{R_{1}}{100})(1+\frac{R_{2}}{100})(1+\frac{R_{3}}{100})\)

=> \(500000(1-\frac{4}{100})(1+\frac{5}{100})(1+\frac{10}{100})\)

= 500,000 (0.96) (1.05) (1.10) = 554,400

Thus, the net profit is Rs 554,400.

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