Each of A, B, C and D need a unique time to do a certain work. A can do the work in x days and B can do the work in 2x days. A started the work and do it for `22(2)/(9)`days then he is replaced by B, who completed remaining work in same time as C and D together can complete the whole work. The ratio of the efficiency of C and D is 4 : 5. If C and D work for alternative days starting from C then they can do the total work in `44(1)/(2)`days.
A and B together can compete 225% of the work in how many days?

Options

A. `66(2)/(3)` days
B. 60 days
C. 50 days
D. 25 days

Answer

To solve the problem step by step, we will follow the information given in the question and derive the necessary values.

### Step 1: Determine the total work done by C and D
- The efficiency ratio of C and D is given as 4:5.
- When C and D work alternately starting from C, they complete the work in \(44 \frac{1}{2}\) days.
- First, convert \(44 \frac{1}{2}\) days into improper fraction:
\[ 44 \frac{1}{2} = \frac{89}{2} \text{ days} \]
- In 2 days, C and D together complete:
\[ 4 + 5 = 9 \text{ units of work} \]
- In \(44\) days, they will complete:
\[ \frac{89}{2} \div 2 = 44.5 \text{ days} \Rightarrow 44 \text{ complete cycles of 2 days} + 1 \text{ day of C} \]
- Therefore, the total work done in 44 complete cycles:
\[ 44 \times 9 = 396 \text{ units} \]
- On the last day, C works for half a day, completing:
\[ \frac{4}{2} = 2 \text{ units} \]
- Thus, the total work done is:
\[ 396 + 2 = 398 \text{ units} \]

### Step 2: Calculate the efficiencies of A and B
- Let A's time to complete the work be \(x\) days, then B's time is \(2x\) days.
- The efficiency of A is:
\[ \text{Efficiency of A} = \frac{1}{x} \text{ (work per day)} \]
- The efficiency of B is:
\[ \text{Efficiency of B} = \frac{1}{2x} \]

### Step 3: Calculate the work done by A in \(22 \frac{2}{9}\) days
- Convert \(22 \frac{2}{9}\) into improper fraction:
\[ 22 \frac{2}{9} = \frac{200}{9} \text{ days} \]
- Work done by A in this time:
\[ \text{Work by A} = \frac{200}{9} \times \frac{1}{x} = \frac{200}{9x} \]

### Step 4: Calculate remaining work
- Total work is \(398\) units, so remaining work after A's contribution is:
\[ \text{Remaining Work} = 398 - \frac{200}{9x} \]

### Step 5: Determine the time taken by B to finish the remaining work
- Since B completes the remaining work in the same time as C and D can complete the whole work, we know:
\[ \frac{398 - \frac{200}{9x}}{\frac{1}{2x}} = \text{Time taken by B} \]
- The time taken by C and D together to complete the whole work is:
\[ \frac{398}{\frac{9}{2}} = \frac{398 \times 2}{9} = \frac{796}{9} \text{ days} \]

### Step 6: Set up the equation
- Equating the time taken by B to finish the remaining work to the time taken by C and D:
\[ \frac{398 - \frac{200}{9x}}{\frac{1}{2x}} = \frac{796}{9} \]
- Solving this equation will yield the value of \(x\).

### Step 7: Calculate the total work A and B can complete together in 225%
- Total work is \(398\) units, so \(225\%\) of the work is:
\[ 398 \times \frac{225}{100} = 448.5 \text{ units} \]
- Combined efficiency of A and B:
\[ \text{Combined Efficiency} = \frac{1}{x} + \frac{1}{2x} = \frac{3}{2x} \]
- Time taken by A and B together to complete \(448.5\) units:
\[ \text{Time} = \frac{448.5}{\frac{3}{2x}} = \frac{448.5 \times 2x}{3} = \frac{897x}{3} \]

### Step 8: Substitute \(x\) and calculate the final answer
- Substitute the value of \(x\) obtained from previous calculations to find the total time.