A certain sum becomes X 600 in a certain time at the rate of 6% simple interest. The same sum amounts to X 200 at the rate of 1% simple interest in the same duration. Find the sum and time.

Options

A. `₹120and66(2)/(3)yr`
B. `₹150and66(2)/(3)yr`
C. `₹130and66(2)/(3)yr`
D. `₹160and66(2)/(3)yr`

Answer

To solve the problem step by step, we will use the information provided about the two cases of simple interest.

### Step 1: Define the variables
Let:
- \( P \) = Principal amount (the sum we need to find)
- \( T \) = Time (the duration for which the interest is calculated)
- \( R_1 = 6\% \) (rate of interest for the first case)
- \( R_2 = 1\% \) (rate of interest for the second case)
- \( A_1 = 600 \) (amount after interest for the first case)
- \( A_2 = 200 \) (amount after interest for the second case)

### Step 2: Use the formula for the amount in simple interest
The formula for the amount in simple interest is given by:
\[ A = P + \frac{P \times R \times T}{100} \]
This can be rearranged to find the principal:
\[ P = \frac{A \times 100}{100 + R \times T} \]

### Step 3: Set up equations for both cases
For the first case:
\[ 600 = P + \frac{P \times 6 \times T}{100} \]
This simplifies to:
\[ 600 = P \left(1 + \frac{6T}{100}\right) \]
For the second case:
\[ 200 = P + \frac{P \times 1 \times T}{100} \]
This simplifies to:
\[ 200 = P \left(1 + \frac{T}{100}\right) \]

### Step 4: Express both equations in terms of \( P \)
From the first case:
\[ P = \frac{600}{1 + \frac{6T}{100}} \quad (1) \]
From the second case:
\[ P = \frac{200}{1 + \frac{T}{100}} \quad (2) \]

### Step 5: Set the two expressions for \( P \) equal to each other
\[ \frac{600}{1 + \frac{6T}{100}} = \frac{200}{1 + \frac{T}{100}} \]

### Step 6: Cross-multiply to eliminate the fractions
\[ 600 \left(1 + \frac{T}{100}\right) = 200 \left(1 + \frac{6T}{100}\right) \]

### Step 7: Expand both sides
\[ 600 + 6T \cdot 6 = 200 + 12T \]
\[ 600 + 6T = 200 + 12T \]

### Step 8: Rearrange to solve for \( T \)
\[ 600 - 200 = 12T - 6T \]
\[ 400 = 6T \]
\[ T = \frac{400}{6} = \frac{200}{3} \text{ years} \]

### Step 9: Substitute \( T \) back to find \( P \)
Using equation (1):
\[ P = \frac{600}{1 + \frac{6 \times \frac{200}{3}}{100}} = \frac{600}{1 + \frac{1200}{300}} = \frac{600}{1 + 4} = \frac{600}{5} = 120 \]

### Final Answer
The principal amount \( P \) is \( 120 \) and the time \( T \) is \( \frac{200}{3} \) years.

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