Working efficiency of A is 20% more than that of B, who can complete a work 'X' in 36 days.
B and C together started to complete the work 'X' and after 10 days they both left the work and then remaining work is done by A alone in 15 days.
A and C together started to complete another work 'Y' and after working for 12 days they both left the work and Remaining work is done by B alone in 16 days. If D first completed work 'X' and then completed work 'Y' in total 38 days.
It is given that efficiency of all, in completing work 'X' and work 'Y' is same.
A, B and C working together completed `(1)/(3)`rd of work 'X', and then A and C are replaced by D. Now remaining of work 'X' is completed by B and D together. For how many days B worked?

Options

A. 12 days
B. 10 days
C. 15 days
D. 4 days

Answer

To solve the problem step by step, let's break it down systematically.

### Step 1: Determine the efficiencies of A, B, C, and D
- Let the efficiency of B be \( x \).
- Since A's efficiency is 20% more than B's, we have:
\[ A = x + 0.2x = 1.2x \]
- Given that B can complete work X in 36 days, the total work \( W \) can be calculated as:
\[ W = \text{Efficiency} \times \text{Time} = x \times 36 \]
- Therefore, the total work \( W = 36x \).

### Step 2: Calculate the total work in terms of B's efficiency
- From the information given, we can also express the total work as:
\[ W = 180 \quad \text{(as derived from the video)} \]
- Setting these equal gives:
\[ 36x = 180 \implies x = 5 \]
- Thus, the efficiencies are:
\[ B = 5, \quad A = 1.2 \times 5 = 6, \quad \text{and we need to find C's efficiency.} \]

### Step 3: Find C's efficiency
- From the work done by B and C together for 10 days:
\[ \text{Work done by B and C in 10 days} = 10(B + C) = 10(5 + C) \]
- The remaining work is completed by A in 15 days:
\[ \text{Remaining work} = 15A = 15 \times 6 = 90 \]
- Therefore, the total work can also be expressed as:
\[ 10(5 + C) + 90 = 180 \]
\[ 50 + 10C + 90 = 180 \implies 10C = 40 \implies C = 4 \]

### Step 4: Efficiency of D
- A and C work together for 12 days, and then B completes the remaining work in 16 days.
- The total work done by A and C in 12 days:
\[ 12(A + C) = 12(6 + 4) = 120 \]
- The remaining work done by B alone in 16 days:
\[ 16B = 16 \times 5 = 80 \]
- Therefore, the total work for Y is:
\[ 120 + 80 = 200 \]

### Step 5: Calculate D's efficiency
- D completes both works in 38 days:
\[ \frac{180}{D} + \frac{200}{D} = 38 \implies \frac{380}{D} = 38 \implies D = 10 \]

### Step 6: Work done by A, B, and C together
- A, B, and C together complete \( \frac{1}{3} \) of work X:
\[ \text{Work done} = \frac{1}{3} \times 180 = 60 \]
- Their combined efficiency:
\[ A + B + C = 6 + 5 + 4 = 15 \]
- Time taken to complete 60 work:
\[ \text{Time} = \frac{60}{15} = 4 \text{ days} \]

### Step 7: Remaining work done by B and D
- Remaining work after A, B, and C completed \( \frac{1}{3} \):
\[ \text{Remaining work} = 180 - 60 = 120 \]
- B and D together work on the remaining work:
\[ \text{Efficiency of B and D} = 5 + 10 = 15 \]
- Time taken to complete the remaining work:
\[ \text{Time} = \frac{120}{15} = 8 \text{ days} \]

### Step 8: Total days worked by B
- B worked for:
\[ 4 \text{ days (with A and C)} + 8 \text{ days (with D)} = 12 \text{ days} \]

### Final Answer
B worked for **12 days**.