Answer :
Let's systematically work through the problem step-by-step:
Given:
1. Ganesh purchased industrial land at Rs. [tex]$1,60,00,000$[/tex].
2. He also purchased a printing machine at Rs. [tex]$5,40,00,000$[/tex].
3. The value of the land increases at a rate of [tex]$20\%$[/tex] per annum (compounded annually).
4. The value of the machine decreases at a rate of [tex]$20\%$[/tex] per annum (compounded annually).
Part (a): Define compound growth of price.
The compound growth of price [tex]\( P_G \)[/tex] is the increase in value over time, which can be described by the formula:
[tex]\[ P_G = P \left(1 + \frac{R}{100}\right)^T - P \][/tex]
Similarly, depreciation [tex]\( P_D \)[/tex] for a decreasing value over time is given by:
[tex]\[ P_D = P - P \left(1 - \frac{R}{100}\right)^T \][/tex]
Part (b): What will be the value of the land after 2 years?
To find the value of the land after [tex]\( T = 2 \)[/tex] years with an initial value of Rs. [tex]$1,60,00,000$[/tex] and an annual growth rate of [tex]$20\%$[/tex], we use the compound interest formula:
[tex]\[ V_{\text{land}} = P \left(1 + \frac{R}{100}\right)^T \][/tex]
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \left(1 + \frac{20}{100}\right)^2 \][/tex]
Calculating it, we get:
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \left(1.2\right)^2 \][/tex]
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \times 1.44 \][/tex]
[tex]\[ V_{\text{land\_2years}} = 2,30,40,000 \][/tex]
So, the value of the land after 2 years will be Rs. [tex]$2,30,40,000$[/tex].
Part (c): What will be the value of the machine after 2 years?
To find the value of the machine after [tex]\( T = 2 \)[/tex] years with an initial value of Rs. [tex]$5,40,00,000$[/tex] and an annual depreciation rate of [tex]$20\%$[/tex], we use the depreciation formula:
[tex]\[ V_{\text{machine}} = P \left(1 - \frac{R}{100}\right)^T \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \left(1 - \frac{20}{100}\right)^2 \][/tex]
Calculating it, we get:
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \left(0.8\right)^2 \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \times 0.64 \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 3,45,60,000 \][/tex]
So, the value of the machine after 2 years will be Rs. [tex]$3,45,60,000$[/tex].
Part (d): In how many years will the value of the machine and the value of the land be equal?
To find the time [tex]\( T \)[/tex] where the values are equal, we set up the equation:
[tex]\[ 1,60,00,000 \left(1 + \frac{20}{100}\right)^T = 5,40,00,000 \left(1 - \frac{20}{100}\right)^T \][/tex]
Simplifying, we get:
[tex]\[ \left(1.2\right)^T = \frac{5,40,00,000}{1,60,00,000} \left(0.8\right)^T \][/tex]
[tex]\[ \left(1.2\right)^T = 3.375 \left(0.8\right)^T \][/tex]
Using logarithms to solve for [tex]\( T \)[/tex]:
[tex]\[ \log \left(1.2^T \right) = \log \left(3.375 \left(0.8\right)^T \right) \][/tex]
[tex]\[ T \cdot \log (1.2) = \log (3.375) + T \cdot \log (0.8) \][/tex]
[tex]\[ T \cdot (\log (1.2) - \log (0.8)) = \log (3.375) \][/tex]
[tex]\[ T = \frac{\log (3.375)}{\log (1.2) - \log (0.8)} \][/tex]
Solving this, we find:
[tex]\[ T = 3 \][/tex]
So, the value of the land and the machine will be equal in 3 years.
Summary of Answers:
(a) Compound growth of price is defined using the formula [tex]\( P_G = P (1 + \frac{R}{100})^T - P \)[/tex] for growth and [tex]\( P_D = P - P (1 - \frac{R}{100})^T \)[/tex] for depreciation.
(b) The value of the land after 2 years will be Rs. [tex]$2,30,40,000$[/tex].
(c) The value of the machine after 2 years will be Rs. [tex]$3,45,60,000$[/tex].
(d) The value of the land and the machine will be equal in 3 years.
Given:
1. Ganesh purchased industrial land at Rs. [tex]$1,60,00,000$[/tex].
2. He also purchased a printing machine at Rs. [tex]$5,40,00,000$[/tex].
3. The value of the land increases at a rate of [tex]$20\%$[/tex] per annum (compounded annually).
4. The value of the machine decreases at a rate of [tex]$20\%$[/tex] per annum (compounded annually).
Part (a): Define compound growth of price.
The compound growth of price [tex]\( P_G \)[/tex] is the increase in value over time, which can be described by the formula:
[tex]\[ P_G = P \left(1 + \frac{R}{100}\right)^T - P \][/tex]
Similarly, depreciation [tex]\( P_D \)[/tex] for a decreasing value over time is given by:
[tex]\[ P_D = P - P \left(1 - \frac{R}{100}\right)^T \][/tex]
Part (b): What will be the value of the land after 2 years?
To find the value of the land after [tex]\( T = 2 \)[/tex] years with an initial value of Rs. [tex]$1,60,00,000$[/tex] and an annual growth rate of [tex]$20\%$[/tex], we use the compound interest formula:
[tex]\[ V_{\text{land}} = P \left(1 + \frac{R}{100}\right)^T \][/tex]
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \left(1 + \frac{20}{100}\right)^2 \][/tex]
Calculating it, we get:
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \left(1.2\right)^2 \][/tex]
[tex]\[ V_{\text{land\_2years}} = 1,60,00,000 \times 1.44 \][/tex]
[tex]\[ V_{\text{land\_2years}} = 2,30,40,000 \][/tex]
So, the value of the land after 2 years will be Rs. [tex]$2,30,40,000$[/tex].
Part (c): What will be the value of the machine after 2 years?
To find the value of the machine after [tex]\( T = 2 \)[/tex] years with an initial value of Rs. [tex]$5,40,00,000$[/tex] and an annual depreciation rate of [tex]$20\%$[/tex], we use the depreciation formula:
[tex]\[ V_{\text{machine}} = P \left(1 - \frac{R}{100}\right)^T \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \left(1 - \frac{20}{100}\right)^2 \][/tex]
Calculating it, we get:
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \left(0.8\right)^2 \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 5,40,00,000 \times 0.64 \][/tex]
[tex]\[ V_{\text{machine\_2years}} = 3,45,60,000 \][/tex]
So, the value of the machine after 2 years will be Rs. [tex]$3,45,60,000$[/tex].
Part (d): In how many years will the value of the machine and the value of the land be equal?
To find the time [tex]\( T \)[/tex] where the values are equal, we set up the equation:
[tex]\[ 1,60,00,000 \left(1 + \frac{20}{100}\right)^T = 5,40,00,000 \left(1 - \frac{20}{100}\right)^T \][/tex]
Simplifying, we get:
[tex]\[ \left(1.2\right)^T = \frac{5,40,00,000}{1,60,00,000} \left(0.8\right)^T \][/tex]
[tex]\[ \left(1.2\right)^T = 3.375 \left(0.8\right)^T \][/tex]
Using logarithms to solve for [tex]\( T \)[/tex]:
[tex]\[ \log \left(1.2^T \right) = \log \left(3.375 \left(0.8\right)^T \right) \][/tex]
[tex]\[ T \cdot \log (1.2) = \log (3.375) + T \cdot \log (0.8) \][/tex]
[tex]\[ T \cdot (\log (1.2) - \log (0.8)) = \log (3.375) \][/tex]
[tex]\[ T = \frac{\log (3.375)}{\log (1.2) - \log (0.8)} \][/tex]
Solving this, we find:
[tex]\[ T = 3 \][/tex]
So, the value of the land and the machine will be equal in 3 years.
Summary of Answers:
(a) Compound growth of price is defined using the formula [tex]\( P_G = P (1 + \frac{R}{100})^T - P \)[/tex] for growth and [tex]\( P_D = P - P (1 - \frac{R}{100})^T \)[/tex] for depreciation.
(b) The value of the land after 2 years will be Rs. [tex]$2,30,40,000$[/tex].
(c) The value of the machine after 2 years will be Rs. [tex]$3,45,60,000$[/tex].
(d) The value of the land and the machine will be equal in 3 years.
