A sum of Rs. P was invested at 10% for 2 years at simple interest. If the same sum was invested at 20% for 'X' years, it would have fetched Rs. 200 more. Find 'x' if Px = 5000. (value of x is given in months)

Options

A. 12
B. 18
C. 15
D. None of these

Answer

To solve the problem step by step, we need to find the value of 'X' in months based on the given conditions. Let's break it down:

### Step 1: Understand the problem
We have a principal amount \( P \) invested at 10% for 2 years, and we need to find out how long (in months) the same amount \( P \) would need to be invested at 20% to earn Rs. 200 more than the interest earned at 10%.

### Step 2: Calculate the simple interest at 10%
The formula for simple interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
For the first investment:
- Rate \( R = 10\% \)
- Time \( T = 2 \) years

Substituting the values:
\[ SI_1 = \frac{P \times 10 \times 2}{100} = \frac{20P}{100} = \frac{P}{5} \]

### Step 3: Calculate the simple interest at 20%
For the second investment:
- Rate \( R = 20\% \)
- Time \( T = X \) years

Substituting the values:
\[ SI_2 = \frac{P \times 20 \times X}{100} = \frac{20PX}{100} = \frac{PX}{5} \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the interest earned at 20% for \( X \) years is Rs. 200 more than the interest earned at 10% for 2 years. Therefore, we can write:
\[ SI_2 - SI_1 = 200 \]
Substituting the values we calculated:
\[ \frac{PX}{5} - \frac{P}{5} = 200 \]

### Step 5: Simplify the equation
Factoring out \( \frac{P}{5} \):
\[ \frac{P(X - 1)}{5} = 200 \]
Multiplying both sides by 5:
\[ P(X - 1) = 1000 \]

### Step 6: Substitute the value of \( PX \)
We know from the problem statement that \( PX = 5000 \). Therefore, we can express \( P \) in terms of \( X \):
\[ P = \frac{5000}{X} \]

### Step 7: Substitute \( P \) back into the equation
Substituting \( P \) in the equation \( P(X - 1) = 1000 \):
\[ \frac{5000}{X}(X - 1) = 1000 \]

### Step 8: Solve for \( X \)
Cross-multiplying gives:
\[ 5000(X - 1) = 1000X \]
Expanding this:
\[ 5000X - 5000 = 1000X \]
Bringing all terms involving \( X \) to one side:
\[ 5000X - 1000X = 5000 \]
\[ 4900X = 5000 \]
Dividing both sides by 4900:
\[ X = \frac{5000}{4900} = \frac{500}{490} = \frac{50}{49} \text{ years} \]

### Step 9: Convert \( X \) to months
To convert years to months, multiply by 12:
\[ X \text{ (in months)} = \frac{50}{49} \times 12 = \frac{600}{49} \approx 12.24 \text{ months} \]

### Step 10: Conclusion
Since the options provided are whole numbers, we can conclude that the closest whole number is 12 months. However, since the question asks for the exact value and does not match any of the options, the answer would be "none of these".