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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.5 PAGE NO: 14.31
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:
Given details are,
Price of a boat is = Rs 16000
Depreciation rate = 5% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Value of the boat after two years is Rs 14440
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:
Given details are,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Total depreciation during this period is Rs (100000 - 81000) = Rs 19000
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:
Given details are,
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.5 PAGE NO: 14.31
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:
Given details are,
Price of a boat is = Rs 16000
Depreciation rate = 5% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Value of the boat after two years is Rs 14440
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:
Given details are,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Total depreciation during this period is Rs (100000 - 81000) = Rs 19000
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:
Given details are,
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Price of land is = Rs 640000
Rate of increase = 5% in every six month
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the plot after two years will be Rs 706440.25
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:
Given details are,
Price of house is = Rs 30000
Depreciation rate is = 25% per year
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the house after 3 years is Rs 12656.25
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:
Given details are,
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)
n
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
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.5 PAGE NO: 14.31
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:
Given details are,
Price of a boat is = Rs 16000
Depreciation rate = 5% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Value of the boat after two years is Rs 14440
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:
Given details are,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Total depreciation during this period is Rs (100000 - 81000) = Rs 19000
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:
Given details are,
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Price of land is = Rs 640000
Rate of increase = 5% in every six month
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the plot after two years will be Rs 706440.25
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:
Given details are,
Price of house is = Rs 30000
Depreciation rate is = 25% per year
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the house after 3 years is Rs 12656.25
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:
Given details are,
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)
n
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
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
? The purchase price is Rs 60000
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:
Given details are,
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)
n
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
? The refrigerator was purchased for Rs 12500
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:
Given details are,
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)
n
= 17000 (1 + 5/100) (1 – 4/100)
= 17000 (105/100) (96/100)
= 17000 (1.05) (0.96)
= 17136
? The cost of TV set in the year 2001 is Rs 17136
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
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.5 PAGE NO: 14.31
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:
Given details are,
Price of a boat is = Rs 16000
Depreciation rate = 5% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Value of the boat after two years is Rs 14440
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:
Given details are,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
By using the formula,
A = P (1 + R/100)
= 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
? Total depreciation during this period is Rs (100000 - 81000) = Rs 19000
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:
Given details are,
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Price of land is = Rs 640000
Rate of increase = 5% in every six month
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the plot after two years will be Rs 706440.25
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:
Given details are,
Price of house is = Rs 30000
Depreciation rate is = 25% per year
By using the formula,
A = P (1 + R/100)
n
= 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
? The value of the house after 3 years is Rs 12656.25
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:
Given details are,
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)
n
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
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
? The purchase price is Rs 60000
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:
Given details are,
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)
n
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
? The refrigerator was purchased for Rs 12500
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:
Given details are,
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)
n
= 17000 (1 + 5/100) (1 – 4/100)
= 17000 (105/100) (96/100)
= 17000 (1.05) (0.96)
= 17136
? The cost of TV set in the year 2001 is Rs 17136
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
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
3 years.
Solution:
Given,
Initial investment by Ashish is = Rs 500000
Incurred loss in the first year is = 4%
Profit in 2
nd
year is = 5 %
Profit in 3
rd
year is = 10%
By using the formula,
A = P (1 + R/100)
n
= 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
? The net profit for the entire period of 3 years is Rs 554400
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