A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs 180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs 120 only. For how many years was the sum invested ?

Options

A. 15 years
B. 20 years
C. 22 years
D. 25 years

Answer

To solve the problem step by step, we will use the formula for Simple Interest (SI):

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Where:
- \( P \) = Principal amount (the initial sum of money)
- \( R \) = Rate of interest per annum
- \( T \) = Time in years

### Step 1: Set up the equations based on the information given.

1. When the sum is invested at 8% per annum, it grows to Rs 180.
\[ \text{SI}_1 = \frac{P \times 8 \times T}{100} \]
Therefore, we can write:
\[ P + \frac{P \times 8 \times T}{100} = 180 \]
This simplifies to:
\[ P \left(1 + \frac{8T}{100}\right) = 180 \quad \text{(Equation 1)} \]

2. When the same sum is invested at 4% per annum, it grows to Rs 120.
\[ \text{SI}_2 = \frac{P \times 4 \times T}{100} \]
Therefore, we can write:
\[ P + \frac{P \times 4 \times T}{100} = 120 \]
This simplifies to:
\[ P \left(1 + \frac{4T}{100}\right) = 120 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations to find \( P \) and \( T \).

From Equation 1:
\[ P \left(1 + \frac{8T}{100}\right) = 180 \]
From Equation 2:
\[ P \left(1 + \frac{4T}{100}\right) = 120 \]

### Step 3: Divide Equation 1 by Equation 2.

\[ \frac{P \left(1 + \frac{8T}{100}\right)}{P \left(1 + \frac{4T}{100}\right)} = \frac{180}{120} \]
This simplifies to:
\[ \frac{1 + \frac{8T}{100}}{1 + \frac{4T}{100}} = \frac{3}{2} \]

### Step 4: Cross-multiply to eliminate the fraction.

\[ 2 \left(1 + \frac{8T}{100}\right) = 3 \left(1 + \frac{4T}{100}\right) \]
Expanding both sides:
\[ 2 + \frac{16T}{100} = 3 + \frac{12T}{100} \]

### Step 5: Rearranging the equation to isolate \( T \).

\[ 2 - 3 = \frac{12T}{100} - \frac{16T}{100} \]
\[ -1 = -\frac{4T}{100} \]
Multiplying both sides by -1:
\[ 1 = \frac{4T}{100} \]
Multiplying both sides by 100:
\[ 100 = 4T \]
Dividing by 4:
\[ T = 25 \]

### Conclusion:

The sum was invested for **25 years**.

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