A principal amounts to X 944 in 3 yr and to X 1040 in 5 yr, each sum being invested at the same simple interest. The principal was

Options

A. X800
B. X991
C. X750
D. X900

Answer

To find the principal amount that amounts to ₹944 in 3 years and ₹1040 in 5 years at the same simple interest rate, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables:**
Let the principal amount be \( P \) and the rate of interest be \( R \).

2. **Use the Simple Interest Formula:**
The formula for Simple Interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
where \( T \) is the time in years.

3. **Set Up the Equations:**
- For the amount of ₹944 after 3 years:
\[ 944 = P + \frac{P \times R \times 3}{100} \]
Rearranging gives:
\[ SI_1 = 944 - P \]
- For the amount of ₹1040 after 5 years:
\[ 1040 = P + \frac{P \times R \times 5}{100} \]
Rearranging gives:
\[ SI_2 = 1040 - P \]

4. **Express SI in Terms of P and R:**
From the first equation:
\[ SI_1 = \frac{P \times R \times 3}{100} \implies R = \frac{100 \times (944 - P)}{3P} \]
From the second equation:
\[ SI_2 = \frac{P \times R \times 5}{100} \implies R = \frac{100 \times (1040 - P)}{5P} \]

5. **Set the Two Expressions for R Equal:**
Since both expressions equal \( R \):
\[ \frac{100 \times (944 - P)}{3P} = \frac{100 \times (1040 - P)}{5P} \]

6. **Cancel Out Common Terms:**
Cancel \( 100 \) and \( P \) (assuming \( P \neq 0 \)):
\[ \frac{944 - P}{3} = \frac{1040 - P}{5} \]

7. **Cross Multiply:**
\[ 5(944 - P) = 3(1040 - P) \]

8. **Distribute:**
\[ 4720 - 5P = 3120 - 3P \]

9. **Rearrange the Equation:**
\[ 4720 - 3120 = 5P - 3P \]
\[ 1600 = 2P \]

10. **Solve for P:**
\[ P = \frac{1600}{2} = 800 \]

### Final Answer:
The principal amount is ₹800.

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