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Class 8 RD Sharma Solutions- Chapter 14 Compound Interest – Exercise 14.5
  • Last Updated : 17 Dec, 2020

Question 1. Ms. Cherian purchases a boat for Rs. 16000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.

Solution:

We have,

Price of a boat is = Rs 16000

Depreciation rate = 5% per annum

By using the formula,



A = P (1 + R/100)t

Substituting the values, we have

= P (1 + R/100)2

Since it is depreciation we use P (1 – R/100)n

= 16000 (1 – 5/100) (1 – 5/100)

= 16000 (95/100) (95/100)

= 16000 (0.95) (0.95)

= 14440

Therefore,

Value of the boat after two years is Rs 14440.

Question 2. 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:

We have,

Present value of machine is = Rs 100000

Rate of depreciation = 10% per annum

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

= 100000 (1 – 10/100) (1 – 10/100)



= 100000 (90/100) (90/100)

= 100000 (0.9) (0.9)

= 81000

Value of machine after two years will be Rs 81000

Therefore,

Total depreciation during this period is Rs (100000 – 81000) = Rs 19000.

Question 3. 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:

We have,

Price of land is = Rs 640000

Rate of increase = 5% in every six month

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

= 640000 (1 + 5/100) (1 + 5/100) (1 + 5/100) (1 + 5/100)

= 640000 (105/100) (105/100) (105/100) (105/100)

= 640000 (1.025) (1.025) (1.025) (1.025)

= 706440.25

Therefore,

The value of the plot after two years will be Rs 706440.25.

Question 4. 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:

We have,

Price of house is = Rs 30000

Depreciation rate is = 25% per year

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

= 30000 (1 – 25/100) (1- 25/100) (1 – 25/100)

= 30000 (75/100) (75/100) (75/100)

= 30000 (0.75) (0.75) (0.75)

= 12656.25

Therefore,

The value of the house after 3 years is Rs 12656.25

Question 5. 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:

We have,

Present value of machine is = Rs 43740

Depreciation rate of machine is = 10% per annum

Let the purchase price 3 years ago be = Rs x

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

43740 = x (1 – 10/100) (1 – 10/100) (1 – 10/100)

43740 = x (90/100) (90/100) (90/100)

43740 = x (0.9) (0.9) (0.9)

43740 = 0.729x

x = 43740/0.729

= 60000

Therefore,

The purchase price is Rs 60000.

Question 6. 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:

We have,

Present value of refrigerator is = Rs 9680

Depreciation rate is = 12%

Let the price of refrigerator 2 years ago be = Rs x

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

9680 = x (1 – 12/100) (1 – 12/100)

9680 = x (88/100) (88/100)

9680 = x (0.88) (0.88)

9680 = 0.7744x

x = 9680/0.7744

= 12500

Therefore,

The refrigerator was purchased for Rs 12500.

Question 7. 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:

We have,

Cost of T.V at beginning of 1999 is = Rs 17000

Hiked in price in the year 2000 is = 5%

Depreciation rate in the year 2001 is = 4%

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

= 17000 (1 + 5/100) (1 – 4/100)

= 17000 (105/100) (96/100)

= 17000 (1.05) (0.96)

= 17136

Therefore,

The cost of TV set in the year 2001 is Rs 17136.

Question 8. 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:

We have,

Initial investment by Ashish is = Rs 500000

Incurred loss in the first year is = 4%

Profit in 2nd year is = 5 %

Profit in 3rd year is = 10%

By using the formula,

A = P (1 + R/100)t

Substituting the values, we have

= 500000 (1 – 4/100) (1 + 5/100) (1 + 10/100)

= 500000 (96/100) (105/100) (110/100)

= 500000 (0.96) (1.05) (1.1)

= 554400

Therefore,

The net profit for the entire period of 3 years is Rs 554400.

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