A building is made of cost ₹ 2,19,700 at a plot of cost 51200. If the rate of plot is increased by 30% annually and the rate of building is decreased by 20% annually, then the cost of plot and building will be equal after that years

Options

A. 3 year
B. 5 year
C. 7 year
D. 1 year

Answer

To solve the problem step by step, we will use the concepts of appreciation and depreciation.

### Step 1: Understand the Initial Values
- Cost of the building = ₹ 2,19,700
- Cost of the plot = ₹ 51,200

### Step 2: Identify the Rates of Change
- The cost of the plot increases by 30% annually.
- The cost of the building decreases by 20% annually.

### Step 3: Formulate the Equations
1. The future cost of the plot after n years can be calculated using the formula for appreciation:
\[ \text{Future Cost of Plot} = \text{Initial Cost of Plot} \times (1 + \frac{r}{100})^n \]
Where \( r = 30 \% \).
\[ \text{Future Cost of Plot} = 51,200 \times (1 + \frac{30}{100})^n = 51,200 \times (1.3)^n \]

2. The future cost of the building after n years can be calculated using the formula for depreciation:
\[ \text{Future Cost of Building} = \text{Initial Cost of Building} \times (1 - \frac{r}{100})^n \]
Where \( r = 20 \% \).
\[ \text{Future Cost of Building} = 2,19,700 \times (1 - \frac{20}{100})^n = 2,19,700 \times (0.8)^n \]

### Step 4: Set the Future Costs Equal
To find the number of years \( n \) when the costs will be equal:
\[ 51,200 \times (1.3)^n = 2,19,700 \times (0.8)^n \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ \frac{(1.3)^n}{(0.8)^n} = \frac{2,19,700}{51,200} \]
This simplifies to:
\[ \left(\frac{1.3}{0.8}\right)^n = \frac{2,19,700}{51,200} \]

### Step 6: Calculate the Right Side
Calculating the right side:
\[ \frac{2,19,700}{51,200} \approx 4.295 \]

### Step 7: Calculate the Left Side
Calculating the left side:
\[ \frac{1.3}{0.8} = 1.625 \]
Thus, we have:
\[ (1.625)^n = 4.295 \]

### Step 8: Take Logarithm on Both Sides
Taking logarithm on both sides:
\[ n \cdot \log(1.625) = \log(4.295) \]
Solving for \( n \):
\[ n = \frac{\log(4.295)}{\log(1.625)} \]

### Step 9: Calculate the Value of n
Using a calculator:
- \( \log(4.295) \approx 0.632 \)
- \( \log(1.625) \approx 0.210 \)

Thus:
\[ n \approx \frac{0.632}{0.210} \approx 3 \]

### Step 10: Conclusion
The costs of the plot and building will be equal after approximately **3 years**.

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