A and B started business with different investments. At start of 2nd ,3rd and 4th quarter each A,B invested additional amounts of 300 rs and 200 rs each and in 4th quarter amount invested by them became equal .If annual profit was divided in the ratio of 5:6 respectively, What was the investment with which A started business?

Options

A. 600rs.
B. 400rs.
C. 300rs.
D. 800rs.

Answer

To solve the problem step by step, we will denote A's initial investment as \( x \) and B's initial investment as \( y \).

### Step 1: Calculate the investment for each quarter

1. **First Quarter (Months 1-3)**:
- A's investment for 3 months = \( 3x \)
- B's investment for 3 months = \( 3y \)

2. **Second Quarter (Months 4-6)**:
- A's investment = \( 3x + 900 \) (A invests an additional 300 each month for 3 months)
- B's investment = \( 3y + 600 \) (B invests an additional 200 each month for 3 months)

3. **Third Quarter (Months 7-9)**:
- A's investment = \( 3x + 1800 \) (A invests an additional 300 each month for 3 months)
- B's investment = \( 3y + 1200 \) (B invests an additional 200 each month for 3 months)

4. **Fourth Quarter (Months 10-12)**:
- A's investment = \( 3x + 2700 \) (A invests an additional 300 each month for 3 months)
- B's investment = \( 3y + 1800 \) (B invests an additional 200 each month for 3 months)

### Step 2: Set up the equation for equal investments in the fourth quarter

According to the problem, in the fourth quarter, A's and B's investments become equal:
\[ 3x + 2700 = 3y + 1800 \]
Rearranging gives:
\[ 3x - 3y = 1800 - 2700 \]
\[ 3x - 3y = -900 \]
Dividing by 3:
\[ x - y = -300 \quad \text{(Equation 1)} \]

### Step 3: Set up the profit-sharing equation

The profit is divided in the ratio of 5:6. The total investments for the entire year would be proportional to their profits. Therefore, we can set up the equation:
\[ \frac{5}{6} = \frac{(3x + 2700)}{(3y + 1800)} \]

Cross-multiplying gives:
\[ 5(3y + 1800) = 6(3x + 2700) \]
Expanding both sides:
\[ 15y + 9000 = 18x + 16200 \]
Rearranging gives:
\[ 15y - 18x = 16200 - 9000 \]
\[ 15y - 18x = 7200 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations

Now we have two equations:
1. \( x - y = -300 \) (Equation 1)
2. \( 15y - 18x = 7200 \) (Equation 2)

From Equation 1, we can express \( y \):
\[ y = x + 300 \]

Substituting \( y \) in Equation 2:
\[ 15(x + 300) - 18x = 7200 \]
Expanding:
\[ 15x + 4500 - 18x = 7200 \]
Combining like terms:
\[ -3x + 4500 = 7200 \]
Rearranging gives:
\[ -3x = 7200 - 4500 \]
\[ -3x = 2700 \]
Dividing by -3:
\[ x = -900 \]

### Step 5: Calculate the initial investment of A

Since \( x = 900 \), this means A's initial investment was:
\[ \text{Investment of A} = 900 \text{ Rs} \]

### Final Answer

Thus, the investment with which A started the business is **900 Rs**.

---