A is thrice as good a workman as B and therefore is able to finish a job in 80 days less than B. If A and B complete `(5)/(8)`th of the job and then A is replaced by C, the remaining job is done by B and C in 15 days. If A and B complete `(5)/(12)`th of the job and then B is replaced by D the remaining job is done by A and D in 10 days.
The job is completed in the following manner: A and B work on day 1, B and C work on day 2, C and D work on day 3, D and A work on day 4 and so on. How long will it take for the job to be completed in this manner?

Options

A. 30 days
B. 24 days
C. 18 days
D. 15 days

Answer

To solve the problem step by step, we will analyze the information given about the work done by A, B, C, and D, and calculate the total time taken to complete the job.

### Step 1: Determine the efficiencies of A and B
Given that A is thrice as good a workman as B, we can denote:
- Efficiency of B = 1 unit of work per day
- Efficiency of A = 3 units of work per day

### Step 2: Calculate the time taken by A and B
Let the time taken by B to complete the job be \( x \) days. Therefore, the time taken by A is \( x - 80 \) days. Since A is thrice as efficient as B, we can set up the equation based on their efficiencies:

\[ \text{Total work} = \text{Efficiency} \times \text{Time} \]

For A:
\[ \text{Total work} = 3 \times (x - 80) \]

For B:
\[ \text{Total work} = 1 \times x \]

Equating the two expressions for total work:
\[ 3(x - 80) = x \]
\[ 3x - 240 = x \]
\[ 2x = 240 \]
\[ x = 120 \]

So, B takes 120 days to complete the job, and A takes:
\[ 120 - 80 = 40 \text{ days} \]

### Step 3: Calculate the total work
Using B's time:
\[ \text{Total work} = 1 \times 120 = 120 \text{ units} \]

### Step 4: Work done by A and B together
A and B complete \( \frac{5}{8} \) of the job:
\[ \text{Work done} = \frac{5}{8} \times 120 = 75 \text{ units} \]

Remaining work:
\[ 120 - 75 = 45 \text{ units} \]

### Step 5: Work done by B and C
B and C complete the remaining work in 15 days:
Let C's efficiency be \( c \). The total efficiency of B and C is:
\[ 1 + c \]
The work done in 15 days is:
\[ 15(1 + c) = 45 \]
\[ 1 + c = 3 \implies c = 2 \]

### Step 6: Work done by A and D
Next, A and B complete \( \frac{5}{12} \) of the job:
\[ \text{Work done} = \frac{5}{12} \times 120 = 50 \text{ units} \]

Remaining work:
\[ 120 - 50 = 70 \text{ units} \]

A and D complete this remaining work in 10 days:
Let D's efficiency be \( d \). The total efficiency of A and D is:
\[ 3 + d \]
The work done in 10 days is:
\[ 10(3 + d) = 70 \]
\[ 3 + d = 7 \implies d = 4 \]

### Step 7: Summary of efficiencies
- Efficiency of A = 3
- Efficiency of B = 1
- Efficiency of C = 2
- Efficiency of D = 4

### Step 8: Work done in a cycle of 4 days
Now, we calculate the work done in the cycle of 4 days:
1. Day 1: A + B = 3 + 1 = 4 units
2. Day 2: B + C = 1 + 2 = 3 units
3. Day 3: C + D = 2 + 4 = 6 units
4. Day 4: D + A = 4 + 3 = 7 units

Total work done in 4 days:
\[ 4 + 3 + 6 + 7 = 20 \text{ units} \]

### Step 9: Total days to complete the job
Total work = 120 units. The number of complete cycles needed:
\[ \frac{120}{20} = 6 \text{ cycles} \]
Each cycle takes 4 days:
\[ 6 \times 4 = 24 \text{ days} \]

Thus, the total time taken to complete the job is **24 days**.