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. 800
B. 991
C. 750
D. 900

Answer

To solve the problem step by step, we will use the concept of Simple Interest (SI) and the formula for calculating the principal amount.

### Step 1: Understand the given information
We have two amounts:
- Amount after 3 years (A1) = X 944
- Amount after 5 years (A2) = X 1040

### Step 2: Identify the time periods
- Time for A1 (T1) = 3 years
- Time for A2 (T2) = 5 years

### Step 3: Use the formula for Simple Interest
The formula for the amount in terms of principal (P), rate (R), and time (T) is:
\[ A = P + SI \]
Where:
\[ SI = \frac{P \times R \times T}{100} \]

### Step 4: Set up equations for both amounts
From the first amount:
\[ A1 = P + \frac{P \times R \times T1}{100} \]
\[ 944 = P + \frac{P \times R \times 3}{100} \] (Equation 1)

From the second amount:
\[ A2 = P + \frac{P \times R \times T2}{100} \]
\[ 1040 = P + \frac{P \times R \times 5}{100} \] (Equation 2)

### Step 5: Rearranging the equations
Rearranging Equation 1 gives:
\[ 944 - P = \frac{P \times R \times 3}{100} \]
\[ 944 - P = \frac{3PR}{100} \] (1)

Rearranging Equation 2 gives:
\[ 1040 - P = \frac{P \times R \times 5}{100} \]
\[ 1040 - P = \frac{5PR}{100} \] (2)

### Step 6: Eliminate R
From Equation (1):
\[ R = \frac{100(944 - P)}{3P} \]

Substituting R from Equation (1) into Equation (2):
\[ 1040 - P = \frac{5P \times \frac{100(944 - P)}{3P}}{100} \]
\[ 1040 - P = \frac{500(944 - P)}{3} \]

### Step 7: Solve for P
Multiply through by 3 to eliminate the fraction:
\[ 3(1040 - P) = 500(944 - P) \]
\[ 3120 - 3P = 472000 - 500P \]
Rearranging gives:
\[ 500P - 3P = 472000 - 3120 \]
\[ 497P = 468880 \]
\[ P = \frac{468880}{497} \]
\[ P = 800 \]

### Conclusion
The principal amount is **X 800**.