Each of A, B, C and D require a unique time to do a certain work. B requires twice the time A requires to do the work. A started the work and do it for 10 days, then he is replaced by B, who worked for four more days than A worked, after that B also left the work. C and D started working on alternative days starting from C and both completed the remaining work in 30 days. The ratio of the efficiency of C and D is 5 : 3 and both together did 32% of the total work.
E and F together work for 12 days, then they are replaced by A and B and they completed the remaining work in 10 days. If the ratio of efficiencies of E and F is 3 : 2 and E and F together completed the whole work then find the difference between the part of work done by E and the part of work done by F in one day?

Options

A. A)`(1)/(200)`
B. B)`(1)/(150)`
C. C)`(1)/(300)`
D. D)`(1)/(75)`

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Understanding the Work Distribution
- Let A's time to complete the work be \( x \) days.
- Then, B's time to complete the work is \( 2x \) days (since B requires twice the time of A).
- The efficiencies of A and B can be expressed as:
- Efficiency of A = \( \frac{1}{x} \)
- Efficiency of B = \( \frac{1}{2x} \)

### Step 2: Work Done by A and B
- A works for 10 days, so the work done by A in 10 days is:
\[ \text{Work by A} = 10 \times \frac{1}{x} = \frac{10}{x} \]
- B works for \( 10 + 4 = 14 \) days (4 more days than A), so the work done by B is:
\[ \text{Work by B} = 14 \times \frac{1}{2x} = \frac{14}{2x} = \frac{7}{x} \]

### Step 3: Total Work Done by A and B
- Total work done by A and B together:
\[ \text{Total Work by A and B} = \frac{10}{x} + \frac{7}{x} = \frac{17}{x} \]

### Step 4: Remaining Work
- The total work is represented as 100%. We know that C and D together completed 32% of the work in 30 days. Therefore, the remaining work done by A and B is:
\[ \text{Remaining Work} = 100\% - 32\% = 68\% \]
- This can be expressed in terms of total work \( W \):
\[ \frac{17}{x} = 0.68W \]

### Step 5: Finding Total Work \( W \)
- Given that C and D together completed 120 units of work (32% of total work):
\[ 0.32W = 120 \implies W = \frac{120}{0.32} = 375 \]

### Step 6: Work Done by A and B
- Substitute \( W \) into the equation:
\[ \frac{17}{x} = 0.68 \times 375 \]
\[ \frac{17}{x} = 255 \implies 17 = 255x \implies x = \frac{17}{255} = \frac{1}{15} \]

### Step 7: Finding Efficiencies
- Now, substituting \( x \) back:
- Efficiency of A = \( \frac{1}{x} = 15 \)
- Efficiency of B = \( \frac{1}{2x} = 7.5 \)

### Step 8: Work Done by E and F
- Let the efficiencies of E and F be \( 3y \) and \( 2y \) respectively.
- They work together for 12 days:
\[ \text{Work by E and F} = (3y + 2y) \times 12 = 5y \times 12 = 60y \]
- After E and F, A and B complete the remaining work in 10 days:
\[ \text{Work by A and B in 10 days} = (15 + 7.5) \times 10 = 225 \]
- Total work done by E and F plus A and B must equal total work:
\[ 60y + 225 = 375 \implies 60y = 150 \implies y = \frac{150}{60} = 2.5 \]

### Step 9: Finding Efficiencies of E and F
- Efficiency of E = \( 3y = 3 \times 2.5 = 7.5 \)
- Efficiency of F = \( 2y = 2 \times 2.5 = 5 \)

### Step 10: Work Done in One Day
- Work done by E in one day:
\[ \text{Work by E in one day} = \frac{7.5}{375} = \frac{1}{50} \]
- Work done by F in one day:
\[ \text{Work by F in one day} = \frac{5}{375} = \frac{1}{75} \]

### Step 11: Finding the Difference
- Difference in work done by E and F in one day:
\[ \text{Difference} = \frac{1}{50} - \frac{1}{75} \]
- Finding a common denominator (150):
\[ = \frac{3}{150} - \frac{2}{150} = \frac{1}{150} \]

### Final Answer
The difference between the part of work done by E and the part of work done by F in one day is:
\[ \frac{1}{150} \]