A man invested Rs. P at `12%` p.a. on simple interest for to years
Quantity I. If at the end of second year he gets Rs. 1200 as interest, then find Rs. P.
Quantity II. Rs. 6000

Options

A. If Quantity I ` gt` Quantity II
B. If Quantity I ` lt` Quantity II
C. If Quantity I ` ge` Quantity II
D. If Quantity I ` le` Quantity II

Answer

To solve the problem, we need to find the principal amount \( P \) invested by the man, given that he received Rs. 1200 as interest at the end of 2 years at a rate of 12% per annum using simple interest.

### Step-by-Step Solution:

1. **Understand the Simple Interest Formula**:
The formula for calculating simple interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
where:
- \( SI \) = Simple Interest
- \( P \) = Principal amount (the initial amount of money)
- \( R \) = Rate of interest per annum
- \( T \) = Time in years

2. **Identify the Given Values**:
From the question:
- \( SI = 1200 \) (the interest earned after 2 years)
- \( R = 12\% \) (the rate of interest)
- \( T = 2 \) years (the time period)

3. **Substitute the Values into the Formula**:
Plugging the values into the simple interest formula:
\[ 1200 = \frac{P \times 12 \times 2}{100} \]

4. **Simplify the Equation**:
Simplifying the right-hand side:
\[ 1200 = \frac{24P}{100} \]
Multiplying both sides by 100 to eliminate the fraction:
\[ 120000 = 24P \]

5. **Solve for \( P \)**:
Now, divide both sides by 24 to find \( P \):
\[ P = \frac{120000}{24} \]
Calculating the division:
\[ P = 5000 \]

6. **Compare with Quantity II**:
Now we compare \( P \) (which is Rs. 5000) with Quantity II (Rs. 6000):
- Quantity I: Rs. 5000
- Quantity II: Rs. 6000

### Conclusion:
Since Rs. 5000 (Quantity I) is less than Rs. 6000 (Quantity II), we conclude that:
- Quantity I < Quantity II

### Final Answer:
The answer is that Quantity I is less than Quantity II.

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