B alone can complete a piece of work in 36 days while D alone can complete same piece of work in 25 days. C is 50 % more efficient than B and A is 20% less efficient than C. If all four starts working together (B & C are working with 56% & 20% more efficiency respectively than their usual efficiency), then in how many days the work will be completed?

Options

A. A)12 days
B. B)18 days
C. C)15 days
D. D)6 days

Answer

To solve the problem step by step, let's break down the information given and calculate the required values.

### Step 1: Determine the efficiency of B and D
- B can complete the work in 36 days.
- D can complete the work in 25 days.

The efficiency of a worker is inversely proportional to the time taken to complete the work. Thus, we can express their efficiencies as follows:

- Efficiency of B (E_B) = 1/36 work/day
- Efficiency of D (E_D) = 1/25 work/day

To make calculations easier, we can convert these efficiencies to a common unit. Let's assume the total work is 3600 units (as calculated later).

- Efficiency of B = 3600/36 = 100 units/day
- Efficiency of D = 3600/25 = 144 units/day

### Step 2: Calculate the efficiency of C
- C is 50% more efficient than B.

Thus, the efficiency of C (E_C) can be calculated as:
\[ E_C = E_B + 0.5 \times E_B = 1.5 \times E_B = 1.5 \times 100 = 150 \text{ units/day} \]

### Step 3: Calculate the efficiency of A
- A is 20% less efficient than C.

Thus, the efficiency of A (E_A) can be calculated as:
\[ E_A = E_C - 0.2 \times E_C = 0.8 \times E_C = 0.8 \times 150 = 120 \text{ units/day} \]

### Step 4: Calculate the modified efficiencies of B and C
- B is working with 56% more efficiency than usual:
\[ \text{Modified efficiency of B} = E_B + 0.56 \times E_B = 1.56 \times E_B = 1.56 \times 100 = 156 \text{ units/day} \]

- C is working with 20% more efficiency than usual:
\[ \text{Modified efficiency of C} = E_C + 0.2 \times E_C = 1.2 \times E_C = 1.2 \times 150 = 180 \text{ units/day} \]

### Step 5: Calculate the total efficiency when A, B, C, and D work together
Now, we can sum the efficiencies of A, B, C, and D:
\[ \text{Total Efficiency} = E_A + \text{Modified efficiency of B} + \text{Modified efficiency of C} + E_D \]
\[ = 120 + 156 + 180 + 144 = 600 \text{ units/day} \]

### Step 6: Calculate the total work
We assumed the total work to be 3600 units.

### Step 7: Calculate the time taken to complete the work
Using the formula:
\[ \text{Time} = \frac{\text{Total Work}}{\text{Total Efficiency}} \]
\[ \text{Time} = \frac{3600}{600} = 6 \text{ days} \]

### Final Answer
The work will be completed in **6 days**.

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