For a certain sum, the simple interest in 2 yr at 8% per annum is 110 more than the simple interest in 1 yr at the rate of 5% per annum for the same sum. Find the sum.

Options

A. X5000
B. X1000
C. X1050
D. X950

Answer

To solve the problem step by step, we will use the formula for simple interest:

**Simple Interest (SI) = (Principal × Rate × Time) / 100**

### Step 1: Define the variables
Let the principal amount (the sum we want to find) be \( P \).

### Step 2: Calculate the simple interest for the first case
For the first case:
- Time \( T_1 = 2 \) years
- Rate \( R_1 = 8\% \)

Using the formula for simple interest:
\[ SI_1 = \frac{P \times R_1 \times T_1}{100} = \frac{P \times 8 \times 2}{100} = \frac{16P}{100} \]

### Step 3: Calculate the simple interest for the second case
For the second case:
- Time \( T_2 = 1 \) year
- Rate \( R_2 = 5\% \)

Using the formula for simple interest:
\[ SI_2 = \frac{P \times R_2 \times T_2}{100} = \frac{P \times 5 \times 1}{100} = \frac{5P}{100} \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the simple interest in the first case is 110 more than the simple interest in the second case:
\[ SI_1 - SI_2 = 110 \]
Substituting the values we calculated:
\[ \frac{16P}{100} - \frac{5P}{100} = 110 \]

### Step 5: Simplify the equation
Combine the terms on the left side:
\[ \frac{16P - 5P}{100} = 110 \]
This simplifies to:
\[ \frac{11P}{100} = 110 \]

### Step 6: Solve for \( P \)
To find \( P \), multiply both sides by 100:
\[ 11P = 110 \times 100 \]
\[ 11P = 11000 \]
Now, divide both sides by 11:
\[ P = \frac{11000}{11} = 1000 \]

### Conclusion
The sum (principal amount) is \( P = 1000 \).

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