A certain sum of money amounts to Rs. 2200 at 5 % p.a. rate of interest, Rs. 2320 at 8 % interest in the same period of time. The peroid of time is :

Options

A. 3 years
B. 4 years
C. 5 years
D. 2 years

Answer

To solve the problem, we need to determine the period of time for which the money is invested based on the given amounts and interest rates. Let's break it down step by step.

### Step 1: Understand the Given Information
We have two amounts:
1. Amount A = Rs. 2200 at a rate of 5% per annum.
2. Amount B = Rs. 2320 at a rate of 8% per annum.

### Step 2: Calculate the Simple Interest for Both Amounts
The formula for the amount (A) in simple interest is:
\[ A = P + SI \]
Where:
- \( A \) = Total amount after time \( t \)
- \( P \) = Principal amount (initial sum of money)
- \( SI \) = Simple Interest

The formula for Simple Interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
Where:
- \( R \) = Rate of interest
- \( T \) = Time in years

### Step 3: Set Up the Equations
For the first amount (Rs. 2200 at 5%):
\[ 2200 = P_1 + \frac{P_1 \times 5 \times T}{100} \]
This simplifies to:
\[ 2200 = P_1 \left(1 + \frac{5T}{100}\right) \]
(1)

For the second amount (Rs. 2320 at 8%):
\[ 2320 = P_2 + \frac{P_2 \times 8 \times T}{100} \]
This simplifies to:
\[ 2320 = P_2 \left(1 + \frac{8T}{100}\right) \]
(2)

### Step 4: Find the Difference Between the Two Amounts
Subtract equation (1) from equation (2):
\[ 2320 - 2200 = P_2 \left(1 + \frac{8T}{100}\right) - P_1 \left(1 + \frac{5T}{100}\right) \]
This gives us:
\[ 120 = P_2 \left(1 + \frac{8T}{100}\right) - P_1 \left(1 + \frac{5T}{100}\right) \]

### Step 5: Express the Principal Amounts in Terms of Time
Let’s assume \( P_1 = P \) and \( P_2 = P \) (the principal amounts are the same for both cases). Then we can express the difference in terms of \( T \):
\[ 120 = P \left(\frac{8T}{100} - \frac{5T}{100}\right) \]
\[ 120 = P \left(\frac{3T}{100}\right) \]

### Step 6: Solve for Time (T)
From the equation:
\[ 120 = \frac{3PT}{100} \]
Rearranging gives:
\[ T = \frac{120 \times 100}{3P} \]
\[ T = \frac{12000}{3P} \]
\[ T = \frac{4000}{P} \]

### Step 7: Substitute the Value of P
From the first amount equation, we can find \( P \):
Using the first equation (1):
Assuming \( P = 2000 \) (as derived from the video transcript):
\[ T = \frac{4000}{2000} = 2 \text{ years} \]

### Conclusion
The period of time is **2 years**.

---