# Class 8 RD Sharma Solutions- Chapter 14 Compound Interest – Exercise 14.5

**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.