A sum was lent out for a certain time. The sum amounts to X 400 at 10% annual interest rate. When the sum was lent out at 4% annual interest rate, it amounts to X 200. Find the sum.

Options

A. `₹(200)/(3)`
B. ₹100
C. `₹(400)/(3)`
D. `₹(500)/(3)`

Answer

To solve the problem step by step, we will use the formula for simple interest and the information provided in the question.

### Step 1: Understand the Problem
We have two scenarios:
1. A sum amounts to ₹400 at a 10% annual interest rate.
2. The same sum amounts to ₹200 at a 4% annual interest rate.

Let’s denote the principal (the sum we want to find) as \( P \) and the time (in years) as \( t \).

### Step 2: Set Up the Equations
Using the formula for the amount in simple interest:
\[ A = P + SI \]
Where \( SI = \frac{P \times R \times t}{100} \).

For the first scenario (10% interest):
\[ 400 = P + \frac{P \times 10 \times t}{100} \]
This simplifies to:
\[ 400 = P + \frac{Pt}{10} \]
Multiplying through by 10 to eliminate the fraction:
\[ 4000 = 10P + Pt \]
So we can write:
\[ 4000 = P(10 + t) \quad \text{(Equation 1)} \]

For the second scenario (4% interest):
\[ 200 = P + \frac{P \times 4 \times t}{100} \]
This simplifies to:
\[ 200 = P + \frac{Pt}{25} \]
Multiplying through by 25:
\[ 5000 = 25P + Pt \]
So we can write:
\[ 5000 = P(25 + t) \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations
Now we have two equations:
1. \( 4000 = P(10 + t) \)
2. \( 5000 = P(25 + t) \)

We can express \( t \) from both equations:
From Equation 1:
\[ t = \frac{4000}{P} - 10 \quad \text{(Equation 3)} \]
From Equation 2:
\[ t = \frac{5000}{P} - 25 \quad \text{(Equation 4)} \]

### Step 4: Set Equations for \( t \) Equal
Setting Equation 3 equal to Equation 4:
\[ \frac{4000}{P} - 10 = \frac{5000}{P} - 25 \]

### Step 5: Solve for \( P \)
Rearranging gives:
\[ \frac{4000}{P} - \frac{5000}{P} = -25 + 10 \]
\[ \frac{-1000}{P} = -15 \]
Multiplying both sides by \( P \):
\[ -1000 = -15P \]
Dividing both sides by -15:
\[ P = \frac{1000}{15} = \frac{200}{3} \]

### Final Answer
Thus, the sum lent out is:
\[ \boxed{\frac{200}{3}} \]