Class 8 Compound Interest 14.5

\begin{array}{l} \text {Q1. Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5 % per annum,} \\ \text {calculate its value after 2 years. } \end{array}\begin{array}{l} V_{n}=V_{0}(1-\frac{R}{100})^{n} \\ \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that Price of boat (} V_{0} \text {) = Rs. 16000.} \\ \text {Rate of depreciation (R) = 5 % per annum} \\ \text {No. of years (n) = 2 years} \\ \text {We know that Value after n years } V_{n}=V_{0}(1-\frac{R}{100})^{n} \\\Rightarrow \quad V_{n}= 16000(1-\frac{5}{100})^{2} \\ =16000(1-\frac{5}{100})^{2} \\ =16000(\frac{95}{100})^{2} \\ = 14440 \\ \therefore \text {Value of boat after 2 years will be Rs. 14440 } \\ \end{array}\begin{array}{l} \text {Q2. The value of a machine depreciates at the rate of 10 % per annum. What will be its value 2 years hence, } \\ \text {if the present value is Rs.100000 ? Also, find the total depreciation during this period.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that Present value of machine (} V_{0} \text {) = Rs. 100000.} \\ \text {Rate of depreciation (R) = 10 % per annum} \\ \text {No. of years (n) = 2 years} \\ \text {We know that Value after n years } V_{n}=V_{0}(1-\frac{R}{100})^{n} \\\Rightarrow \quad V_{n}= 100000(1-\frac{10}{100})^{2} \\ =100000(1-\frac{10}{100})^{2} \\ =100000(\frac{90}{100})^{2} \\ = 81000 \\ \therefore \text {Value of the machine after 2 years will be Rs. 81000 } \\ \text {Total depriciation } =100000 – 81000 = \text {Rs.} 19000 \end{array}\begin{array}{l} \text {Q3. Pritam bought a plot of land for Rs. 640000. Its value is increasing by 5 % of its previous value } \\ \text { after every six months. What will be the value of the plot after 2 years?} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that Present value of plot = Rs. 640000.} \\ \text {Rate of appreciation (R) = 5 % every six month} \\ \text {No. of years = 2 years but increment is happening every 6 months so for 2 yrs n = 4} \\\text {We know that A } =P(1+\frac{R}{100})^{n} \\ \\\Rightarrow \quad A= 640000(1+\frac{5}{100})^{4} \\ =640000(\frac{105}{100})^{4} \\ = 706440.25 \\\text {Hence, The value of plot will be Rs. 706440.25 after 2 yrs.}\end{array}\begin{array}{l} \text {Q4. Mohan purchased a house for Rs. 30000 and its value is depreciating at the rate of 25 % per year. } \\ \text {Find the value of the house after 3 years.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that Mohan purchased a house at (} V_{0} \text {) = Rs. 30000.} \\ \text {Rate of depreciation (R) = 25 % per annum} \\ \text {No. of years (n) = 3 years} \\ \text {We know that Value after n years } V_{n}=V_{0}(1-\frac{R}{100})^{n} \\\Rightarrow \quad V_{n}= 30000(1-\frac{10}{100})^{3} \\ =30000(1-\frac{25}{100})^{3} \\ =30000(\frac{75}{100})^{3} \\ = 12656.25 \\ \therefore \text {Value of the house after 3 years will be Rs. 12656.25 } \\\end{array}\begin{array}{l} \text {Q5. The value of a machine depreciates at the rate of 10 % per annum. It was purchased 3 years ago.} \\ \text {If its present value is Rs. 43740, find its purchase price.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that value of machine 3 yrs ago (} V_{0} \text {) be Rs. x} \\ \text {Rate of depreciation (R) = 10 % per annum} \\ \text {No. of years (n) = 3 years} \\ \text {Present Value of machine } V_{n} = 43740\\\text {We know that Value after n years } V_{n}=V_{0}(1-\frac{R}{100})^{n} \\\Rightarrow \quad 43740= x(1-\frac{10}{100})^{3} \\ \Rightarrow \quad 43740= x(\frac{90}{100})^{3} \\ \Rightarrow \quad x=43740(\frac{100}{90})^{3} \\ \Rightarrow \quad x=60000 \\\therefore \text {The puchased price of the machine was Rs. 60000 } \\ \end{array}\begin{array}{l} \text {Q6. The value of a refrigerator which was purchased 2 years ago, depreciates at 12 % per annum.} \\ \text {If its present value is Rs. 9680, for how much was it purchased? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that value of refrigerator 2 yrs ago (} V_{0} \text {) be Rs. x} \\ \text {Rate of depreciation (R) = 12 % per annum} \\ \text {No. of years (n) = 2 years} \\ \text {Present Value of machine } V_{n} = 9680 \\\text {We know that Value after n years } V_{n}=V_{0}(1-\frac{R}{100})^{n} \\\Rightarrow \quad 9680= x(1-\frac{12}{100})^{2} \\ \Rightarrow \quad 9680= x(\frac{88}{100})^{2} \\ \Rightarrow \quad x=9680(\frac{100}{88})^{2} \\ \Rightarrow \quad x=12500 \\\therefore \text {The puchased price of the refrigerator was Rs. 12500 } \\\end{array}\begin{array}{l} \text {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 %.} \\ \text {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 ? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that inital cost of TV in 1999 was = Rs. 17000.} \\ \text {After that cost got incremented by } R_{1} = 5 \% \text { and decreased by } R_{2} =4 \% \text { respectively. } \\\Rightarrow \quad \text {Cost of the TV } = P(1+\frac{R_{1}}{100})(1-\frac{R_{2}}{100}) \\ \Rightarrow \quad \text {Cost of the TV } = 17000(1+\frac{5}{100})(1-\frac{4}{100}) \\ \Rightarrow \quad \text {Cost of the TV } = 17000(\frac{105}{100})(\frac{96}{100}) \\ \Rightarrow \quad \text {Cost of the TV } = 17136 \\ \therefore \text {The cost of the TV in 2001 was Rs. 17136} \end{array}\begin{array}{l} \text {Q8. Ashish started the business with an initial investment of Rs. 500000. In the first year he incurred a loss of 4 %.} \\ \text {However during the second year he earned a profit of 5 % which in third year rose to 10 %. } \\ \text {Calculate the net profit for the entire period of 3 years.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that inital investment was = Rs. 500000.} \\ \text {First year loss percentage } R_{1} = 4 % \text { and then profit percentage } R_{2} = 5 % \text { and then profit percentage } R_{3} = 10 % \text { respectively. }\Rightarrow \quad \text {Value of business after 3 yrs } = 500000(1-\frac{4}{100})(1+\frac{5}{100})(1+\frac{10}{100})\\ \Rightarrow \quad \text {Value of business after 3 yrs } = 500000(\frac{96}{100})(\frac{105}{100})(\frac{110}{100})\\\Rightarrow \quad \text {Value of business after 3 yrs } = 554400\end{array}