A building is made of cost `Rs. 2,19,700` at a plot of cost `Rs. 51200` . If the rate of plot is increased by `30%` annualy 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. `7` yr
B. `4` yr
C. `3` yr
D. `2` yr

Answer

To solve the problem step by step, we need to calculate the future costs of the plot and the building after 'n' years, considering the percentage increase and decrease.

### Step 1: Define the initial costs
- Cost of the plot (P) = Rs. 51,200
- Cost of the building (B) = Rs. 2,19,700

### Step 2: Determine the annual percentage changes
- The plot's cost increases by 30% annually.
- The building's cost decreases by 20% annually.

### Step 3: Write the formula for the cost of the plot after 'n' years
The future cost of the plot after 'n' years can be calculated using the formula for compound interest:
\[ \text{Cost of plot after } n \text{ years} = P \times (1 + \frac{r}{100})^n \]
Where:
- \( P = 51,200 \)
- \( r = 30 \)

Thus, the formula becomes:
\[ \text{Cost of plot after } n \text{ years} = 51,200 \times (1 + \frac{30}{100})^n = 51,200 \times (1.3)^n \]

### Step 4: Write the formula for the cost of the building after 'n' years
The future cost of the building after 'n' years can also be calculated using the formula for compound interest, but since the cost is decreasing, we use:
\[ \text{Cost of building after } n \text{ years} = B \times (1 - \frac{r}{100})^n \]
Where:
- \( B = 2,19,700 \)
- \( r = 20 \)

Thus, the formula becomes:
\[ \text{Cost of building after } n \text{ years} = 2,19,700 \times (1 - \frac{20}{100})^n = 2,19,700 \times (0.8)^n \]

### Step 5: Set the two equations equal to each other
To find when the costs will be equal:
\[ 51,200 \times (1.3)^n = 2,19,700 \times (0.8)^n \]

### Step 6: Simplify the equation
Dividing both sides by 51,200:
\[ (1.3)^n = \frac{2,19,700}{51,200} \times (0.8)^n \]

Calculating \( \frac{2,19,700}{51,200} \):
\[ \frac{2,19,700}{51,200} \approx 4.295 \]

Thus, we have:
\[ (1.3)^n = 4.295 \times (0.8)^n \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \frac{(1.3)^n}{(0.8)^n} = 4.295 \]
This can be simplified to:
\[ (1.625)^n = 4.295 \]

### Step 8: Taking logarithm on both sides
Taking logarithm:
\[ n \cdot \log(1.625) = \log(4.295) \]

### Step 9: Solve for 'n'
Now, solving for 'n':
\[ n = \frac{\log(4.295)}{\log(1.625)} \]

Calculating the values:
- \( \log(4.295) \approx 0.632 \)
- \( \log(1.625) \approx 0.210 \)

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

### Final Answer
The cost of the plot and the building will be equal after **3 years**.

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