A man takes a loan of 6,000 from his friend on 1st January 2019 with the condition that he will repay with accred simple interest at the rate 6.25%, as and when the interest touches 75. On which date does the loan period expire?

Options

A. 14th February 2019
B. 15th January 2020
C. 20th October 2019
D. 15th March 2019

Answer

To solve the problem step-by-step, we need to calculate the time it takes for the simple interest on the loan to reach Rs 75.

### Step 1: Identify the given values
- Principal (P) = Rs 6,000
- Rate of interest (R) = 6.25%
- Simple Interest (SI) = Rs 75

### Step 2: Use the formula for Simple Interest
The formula for Simple Interest is:
\[ SI = \frac{P \times R \times T}{100} \]
Where:
- SI = Simple Interest
- P = Principal
- R = Rate of interest
- T = Time (in years)

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula:
\[ 75 = \frac{6000 \times 6.25 \times T}{100} \]

### Step 4: Simplify the equation
To simplify, we can multiply both sides by 100:
\[ 7500 = 6000 \times 6.25 \times T \]

### Step 5: Calculate \(6000 \times 6.25\)
Calculating \(6000 \times 6.25\):
\[ 6000 \times 6.25 = 37500 \]
So the equation becomes:
\[ 7500 = 37500 \times T \]

### Step 6: Solve for T
Now, divide both sides by 37500 to find T:
\[ T = \frac{7500}{37500} = \frac{1}{5} \text{ years} \]

### Step 7: Convert years into months
To convert years into months, we multiply by 12 (since there are 12 months in a year):
\[ T = \frac{1}{5} \times 12 = 2.4 \text{ months} \]

### Step 8: Calculate the maturity date
Now, we need to add 2.4 months to the loan date, which is 1st January 2019.
- 2.4 months can be interpreted as 2 months and approximately 12 days (0.4 of a month is about 12 days).

Adding 2 months to 1st January 2019 gives us:
- 1st March 2019

Adding approximately 12 days to 1st March 2019 gives us:
- 13th March 2019

### Final Answer
Thus, the loan period expires on **15th March 2019**.

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