Two equal sums were lent, one at the rate of 11% p.a. for five years and the other at the rate of 8% p.a. for six years, both under simple interest. If the difference in interest accrued in the two cases is 1008. find the sum.

Options

A. 11200
B. 5600
C. 12600
D. 14400

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Define the Variables
Let the sum of money lent be \( S \). Since we are dealing with two equal sums, we can denote the sum as \( S = 100x \).

### Step 2: Calculate Simple Interest for the First Case
For the first case:
- Principal \( P = 100x \)
- Rate \( R = 11\% \)
- Time \( T = 5 \) years

The formula for Simple Interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the values:
\[ SI_1 = \frac{100x \times 11 \times 5}{100} = 55x \]

### Step 3: Calculate Simple Interest for the Second Case
For the second case:
- Principal \( P = 100x \)
- Rate \( R = 8\% \)
- Time \( T = 6 \) years

Using the same formula for Simple Interest:
\[ SI_2 = \frac{100x \times 8 \times 6}{100} = 48x \]

### Step 4: Set Up the Equation for the Difference in Interest
According to the problem, the difference in interest accrued in the two cases is \( 1008 \):
\[ SI_1 - SI_2 = 1008 \]
Substituting the values from our calculations:
\[ 55x - 48x = 1008 \]
This simplifies to:
\[ 7x = 1008 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{1008}{7} = 144 \]

### Step 6: Find the Total Sum
Now that we have \( x \), we can find the total sum \( S \):
\[ S = 100x = 100 \times 144 = 14400 \]

### Final Answer
The sum lent is \( 14400 \).

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