A certain sum of money at simple interest amounts to ₹ 520 in 5 years and ₹ 568 in 7 years. Find the sum of money.

Options

A. Rs 500
B. Rs 800
C. Rs 400
D. Rs 1000

Answer

To solve the problem step by step, we will use the information given about the amounts at different times and the concept of simple interest.

### Step 1: Understand the given information
We know that:
- The amount after 5 years (A1) = ₹520
- The amount after 7 years (A2) = ₹568

### Step 2: Set up the equations
In simple interest, the amount (A) can be expressed as:
\[ A = P + SI \]
where \( P \) is the principal amount and \( SI \) is the simple interest.

The simple interest for a certain number of years can be calculated as:
\[ SI = P \times r \times t \]
where \( r \) is the rate of interest and \( t \) is the time in years.

### Step 3: Write the equations for both amounts
From the information given:
1. For 5 years:
\[ A_1 = P + SI_1 \]
\[ 520 = P + (P \times r \times 5) \]
\[ 520 = P(1 + 5r) \] (Equation 1)

2. For 7 years:
\[ A_2 = P + SI_2 \]
\[ 568 = P + (P \times r \times 7) \]
\[ 568 = P(1 + 7r) \] (Equation 2)

### Step 4: Subtract the two equations
Now, we can subtract Equation 1 from Equation 2 to eliminate \( P \):
\[ 568 - 520 = P(1 + 7r) - P(1 + 5r) \]
\[ 48 = P(7r - 5r) \]
\[ 48 = P(2r) \]

### Step 5: Solve for \( P \)
From the equation \( 48 = P(2r) \), we can express \( P \) as:
\[ P = \frac{48}{2r} \]
\[ P = \frac{24}{r} \] (Equation 3)

### Step 6: Substitute \( P \) back into one of the original equations
Now, we can substitute \( P \) from Equation 3 into either Equation 1 or Equation 2. Let's use Equation 1:
\[ 520 = \frac{24}{r}(1 + 5r) \]
Multiply both sides by \( r \):
\[ 520r = 24(1 + 5r) \]
\[ 520r = 24 + 120r \]
Now, rearranging gives:
\[ 520r - 120r = 24 \]
\[ 400r = 24 \]
\[ r = \frac{24}{400} \]
\[ r = 0.06 \] or \( 6\% \)

### Step 7: Find the principal amount \( P \)
Now, substitute \( r \) back into Equation 3:
\[ P = \frac{24}{0.06} \]
\[ P = 400 \]

### Conclusion
The principal amount is ₹400.