A certain sum of money is invested in two parts at the rate of 1`6(2)/(3)%` per annum compounded annually for 7 years and 10 years respectively. If amount received on both investement is equal. If difference between their investment is ₹ 2540. Then find the total investment.

Options

A. ₹11180
B. ₹ 10062
C. ₹12298
D. ₹ 12700

Answer

To solve the problem step by step, we will follow the details provided in the video transcript and break it down into manageable parts.

### Step 1: Understand the Rate of Interest
The rate of interest given is \( 16 \frac{2}{3} \% \) per annum. We can convert this to a fraction:
\[ 16 \frac{2}{3} \% = \frac{50}{3} \% = \frac{50}{300} = \frac{1}{6} \]

### Step 2: Set Up the Problem
Let the two investments be \( P_A \) and \( P_B \) such that:
- \( P_A \) is invested for 7 years.
- \( P_B \) is invested for 10 years.

According to the problem, the amounts received from both investments are equal:
\[ P_A \left(1 + \frac{1}{6}\right)^7 = P_B \left(1 + \frac{1}{6}\right)^{10} \]

### Step 3: Simplify the Equation
We can express \( 1 + \frac{1}{6} \) as \( \frac{7}{6} \):
\[ P_A \left(\frac{7}{6}\right)^7 = P_B \left(\frac{7}{6}\right)^{10} \]

Dividing both sides by \( \left(\frac{7}{6}\right)^7 \):
\[ P_A = P_B \left(\frac{7}{6}\right)^{3} \]

### Step 4: Express the Difference Between Investments
We know the difference between the two investments is ₹2540:
\[ P_A - P_B = 2540 \]
Substituting \( P_A \) from the previous equation:
\[ P_B \left(\frac{7}{6}\right)^{3} - P_B = 2540 \]

### Step 5: Factor Out \( P_B \)
Factoring out \( P_B \):
\[ P_B \left(\left(\frac{7}{6}\right)^{3} - 1\right) = 2540 \]

### Step 6: Calculate \( \left(\frac{7}{6}\right)^{3} \)
Calculating \( \left(\frac{7}{6}\right)^{3} \):
\[ \left(\frac{7}{6}\right)^{3} = \frac{343}{216} \]
Thus,
\[ \frac{343}{216} - 1 = \frac{343 - 216}{216} = \frac{127}{216} \]

### Step 7: Substitute Back
Now substituting back into the equation:
\[ P_B \left(\frac{127}{216}\right) = 2540 \]
Solving for \( P_B \):
\[ P_B = 2540 \cdot \frac{216}{127} \]

### Step 8: Calculate \( P_B \)
Calculating \( P_B \):
\[ P_B = 2540 \cdot \frac{216}{127} = 2540 \cdot 1.7007874 \approx 4320 \]

### Step 9: Find \( P_A \)
Now substituting \( P_B \) back to find \( P_A \):
\[ P_A = P_B \left(\frac{7}{6}\right)^{3} = 4320 \cdot \frac{343}{216} \approx 6480 \]

### Step 10: Find Total Investment
The total investment \( P_A + P_B \):
\[ Total = P_A + P_B = 6480 + 4320 = 10800 \]

### Final Answer
The total investment is ₹10800.

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