A certain sum was invested by a person on SI at 5% per annum. After 6 months he again made an investment of the same amount on SI at 6% per annum after. A certain period of time the amount received from both invest mentare equal. Which is Rs 4600. Find the sum he has invested in each investment.

Options

A. 4000
B. 4500
C. 3500
D. 5000

Answer

To solve the problem step by step, we need to find the principal amount invested in each investment. Let's denote the principal amount as \( P \).

### Step 1: Understand the investments and their interest calculations
1. The first investment is at 5% per annum for a period of \( t_1 \) years.
2. The second investment is made after 6 months (or 0.5 years) at 6% per annum for a period of \( t_2 \) years.

### Step 2: Set up the equations for the amounts
The total amount received from both investments is Rs 4600.

- For the first investment:
\[ A_1 = P + \text{SI}_1 = P + \left( \frac{P \times 5 \times t_1}{100} \right) \]

- For the second investment:
\[ A_2 = P + \text{SI}_2 = P + \left( \frac{P \times 6 \times t_2}{100} \right) \]

### Step 3: Set the amounts equal to each other
Since both amounts are equal to Rs 4600:
\[ A_1 = A_2 = 4600 \]

### Step 4: Express \( t_2 \) in terms of \( t_1 \)
Since the second investment starts after 6 months, we can express \( t_2 \) as:
\[ t_2 = t_1 - 0.5 \]

### Step 5: Substitute \( t_2 \) in the second amount equation
Now we can write the equations:
1. For the first investment:
\[ 4600 = P + \left( \frac{P \times 5 \times t_1}{100} \right) \]
Simplifying gives:
\[ 4600 = P \left(1 + \frac{5t_1}{100}\right) \]

2. For the second investment:
\[ 4600 = P + \left( \frac{P \times 6 \times (t_1 - 0.5)}{100} \right) \]
Simplifying gives:
\[ 4600 = P \left(1 + \frac{6(t_1 - 0.5)}{100}\right) \]

### Step 6: Set the two equations equal to each other
From the two equations we can set:
\[ P \left(1 + \frac{5t_1}{100}\right) = P \left(1 + \frac{6(t_1 - 0.5)}{100}\right) \]

### Step 7: Cancel \( P \) (assuming \( P \neq 0 \))
\[ 1 + \frac{5t_1}{100} = 1 + \frac{6(t_1 - 0.5)}{100} \]

### Step 8: Solve for \( t_1 \)
Subtracting 1 from both sides:
\[ \frac{5t_1}{100} = \frac{6(t_1 - 0.5)}{100} \]

Multiplying through by 100:
\[ 5t_1 = 6(t_1 - 0.5) \]

Expanding:
\[ 5t_1 = 6t_1 - 3 \]

Rearranging gives:
\[ 6t_1 - 5t_1 = 3 \implies t_1 = 3 \text{ years} \]

### Step 9: Find \( t_2 \)
Using \( t_1 \):
\[ t_2 = t_1 - 0.5 = 3 - 0.5 = 2.5 \text{ years} \]

### Step 10: Substitute \( t_1 \) back to find \( P \)
Using the first investment equation:
\[ 4600 = P \left(1 + \frac{5 \times 3}{100}\right) \]
\[ 4600 = P \left(1 + 0.15\right) = P \times 1.15 \]
\[ P = \frac{4600}{1.15} = 4000 \]

### Final Answer
The sum invested in each investment is Rs 4000.

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