RD Sharma solution class 8 chapter 14 Compound Interest Exercise 14.5

Exercise 14.5

Page-14.5

Question 1:

Ms. Cherian purchased 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.

Answer 1:

$$\begin{gathered} \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{boat} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ \Rightarrow 16,000{\left(1-\frac{5}{100}\right)}^2 \\ =16,000{\left(0.95\right)}^2 \\ =14,440 \\ \mathrm{Thus}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{boat} \mathrm{after} \mathrm{two} \mathrm{years} \mathrm{will} \mathrm{be} \mathrm{Rs} 14,440. \end{gathered}$$

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.

Answer 2:

$$\begin{gathered} \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ \Rightarrow 100,000{\left(1-\frac{10}{100}\right)}^2 \\ =100,000{\left(0.90\right)}^2 \\ =81,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{after} \mathrm{two} \mathrm{years} \mathrm{will} \mathrm{be} \mathrm{Rs} 81,000. \\ \mathrm{Depreciation}=\mathrm{Rs} 100,000-\mathrm{Rs} 81,000 \\ =\mathrm{Rs} 19,000 \end{gathered}$$

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?

Answer 3:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{P}= \mathrm{Rs} 64,000 \\ \mathrm{R}= 5\% \mathrm{for} \mathrm{every} \mathrm{six} \mathrm{months} \\ \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{plot} \mathrm{after} \mathrm{two} \mathrm{years} = \mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ \Rightarrow 64,000{\left(1+\frac{5}{200}\right)}^4 \\ =64,000{\left(1.025\right)}^4 \\ =706,440.25 \\ \mathrm{Thus}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{plot} \mathrm{after} \mathrm{two} \mathrm{years} \mathrm{will} \mathrm{be} \mathrm{Rs} 706,440.25. \end{gathered}$$

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.

Answer 4:

$$\begin{gathered} \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{house} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ \Rightarrow 30,000{\left(1-\frac{25}{100}\right)}^3 \\ =30,000{\left(0.75\right)}^3 \\ =12,656.25 \\ \mathrm{Thus}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{house} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{will} \mathrm{be} \mathrm{Rs} 12,656.25. \end{gathered}$$

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.

Answer 5:

$$\begin{gathered} \mathrm{Purchase} \mathrm{price}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{-\mathrm{n}} \\ \Rightarrow 43,740{\left(1-\frac{10}{100}\right)}^{-3} \\ =43,740{\left(0.90\right)}^{-3} \\ =60,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{purchase} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{was} \mathrm{Rs} 60,000. \end{gathered}$$

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?

Answer 6:

$$\begin{gathered} \mathrm{Purchase} \mathrm{price}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{-\mathrm{n}} \\ \Rightarrow 9,680{\left(1-\frac{12}{100}\right)}^{-2} \\ =9,680{\left(0.88\right)}^{-2} \\ =12,500 \\ \mathrm{Thus}, \mathrm{the} \mathrm{purchase} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{refrigerator} \mathrm{was} \mathrm{Rs} 12,500. \end{gathered}$$

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?

Answer 7:

$$\begin{gathered} \mathrm{Cost} \mathrm{of} \mathrm{the} \mathrm{TV}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)\left(1-\frac{\mathrm{R}}{100}\right) \\ \Rightarrow 17,000\left(1+\frac{5}{100}\right)\left(1-\frac{4}{100}\right) \\ =17,000\left(1.05\right)\left(0.96\right) \\ =17,136 \\ \mathrm{Thus}, \mathrm{the} \mathrm{cost} \mathrm{the} \mathrm{TV} \mathrm{in} 2001 \mathrm{was} \mathrm{Rs} 17,136. \end{gathered}$$

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.

Answer 8:

$$\begin{gathered} \mathrm{Profit} \mathrm{for} \mathrm{three} \mathrm{year}s=\mathrm{P}\left(1-\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ \Rightarrow 500,000\left(1-\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{10}{100}\right) \\ =500,000\left(0.96\right)\left(1.05\right)\left(1.10\right) \\ =554,400 \\ \mathrm{Thus}, \mathrm{the} \mathrm{net} \mathrm{profit} \mathrm{is} \mathrm{Rs} 554,400. \end{gathered}$$