A sum of Rs, 7,930 is divided into three parts and given on loan at 5% simple interset to A, B and C for 2,3 and 4 years respectively. If the amounts of all the three are equal after their respective periods of loan. Then A received a loan of

Options

A. Rs 3,050
B. Rs, 2,760
C. Rs, 2,750
D. Rs. 2,800

Answer

To solve the problem, we need to determine how much A received as a loan from a total sum of Rs. 7,930, which is divided into three parts given to A, B, and C at different time periods but with the same interest rate, such that the amounts received at the end of their respective periods are equal.

### Step-by-Step Solution:

1. **Identify the Variables:**
Let the amounts loaned to A, B, and C be represented as \( A \), \( B \), and \( C \) respectively.

2. **Understand the Interest Calculation:**
The formula for the amount \( A \) after a certain time with simple interest is:
\[ \text{Amount} = \text{Principal} + \text{Interest} = P + \left( P \times \frac{r}{100} \times t \right) \]
where \( P \) is the principal, \( r \) is the rate of interest, and \( t \) is the time in years.

3. **Set Up the Equations:**
For A (2 years):
\[ \text{Amount}_A = A + \left( A \times \frac{5}{100} \times 2 \right) = A \left( 1 + \frac{10}{100} \right) = A \times 1.1 \]
For B (3 years):
\[ \text{Amount}_B = B + \left( B \times \frac{5}{100} \times 3 \right) = B \left( 1 + \frac{15}{100} \right) = B \times 1.15 \]
For C (4 years):
\[ \text{Amount}_C = C + \left( C \times \frac{5}{100} \times 4 \right) = C \left( 1 + \frac{20}{100} \right) = C \times 1.2 \]

4. **Set the Amounts Equal:**
Since all amounts are equal after their respective periods:
\[ A \times 1.1 = B \times 1.15 = C \times 1.2 \]

5. **Express B and C in terms of A:**
From the equality:
\[ B = \frac{A \times 1.1}{1.15} \]
\[ C = \frac{A \times 1.1}{1.2} \]

6. **Substitute into the Total Sum:**
The total sum of the loans is given as Rs. 7,930:
\[ A + B + C = 7930 \]
Substituting for B and C:
\[ A + \frac{A \times 1.1}{1.15} + \frac{A \times 1.1}{1.2} = 7930 \]

7. **Combine the Terms:**
To combine, find a common denominator:
\[ A + \frac{A \times 1.1 \times 1.2 + A \times 1.1 \times 1.15}{1.15 \times 1.2} = 7930 \]
Simplifying the numerator:
\[ A + \frac{A \times 1.1 (1.2 + 1.15)}{1.15 \times 1.2} = 7930 \]
\[ A + \frac{A \times 1.1 \times 2.35}{1.38} = 7930 \]

8. **Factor Out A:**
\[ A \left( 1 + \frac{1.1 \times 2.35}{1.38} \right) = 7930 \]

9. **Calculate the Coefficient:**
Calculate the coefficient and solve for \( A \):
\[ A \left( 1 + \frac{2.585}{1.38} \right) = 7930 \]
\[ A \left( 1 + 1.875 \right) = 7930 \]
\[ A \times 2.875 = 7930 \]
\[ A = \frac{7930}{2.875} \approx 2767 \]

### Final Answer:
A received a loan of **Rs. 2,767**.