Each of A, B, C and D need a umique
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 and B 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 form C then they can do the total work
in `44 1/2` days.

Options

A. 66(2/3)
B. 33(1/3)
C. 16(2/3)
D. 14(2/7)

Explanation

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

### Step 1: Understand the Work Done by A
A can complete the work in \( x \) days. Therefore, A's work rate (efficiency) is:
\[ \text{Efficiency of A} = \frac{1}{x} \text{ (work per day)} \]

### Step 2: Calculate Work Done by A in 22 and \( \frac{2}{9} \) Days
A works for \( 22 \frac{2}{9} \) days. We convert this to an improper fraction:
\[ 22 \frac{2}{9} = \frac{200}{9} \text{ days} \]
The work done by A in this time is:
\[ \text{Work done by A} = \text{Efficiency of A} \times \text{Time} = \frac{1}{x} \times \frac{200}{9} = \frac{200}{9x} \]

### Step 3: Remaining Work
Since the total work is 1 (whole work), the remaining work after A's contribution is:
\[ \text{Remaining Work} = 1 - \frac{200}{9x} = \frac{9x - 200}{9x} \]

### Step 4: Work Done by B
B can complete the work in \( 2x \) days, so B's efficiency is:
\[ \text{Efficiency of B} = \frac{1}{2x} \]
Let \( t \) be the time taken by B to complete the remaining work. The work done by B can be expressed as:
\[ \text{Work done by B} = \text{Efficiency of B} \times t = \frac{1}{2x} \times t \]
Setting this equal to the remaining work:
\[ \frac{1}{2x} \times t = \frac{9x - 200}{9x} \]

### Step 5: Time Taken by C and D Together
It is given that B completes the remaining work in the same time as C and D together can complete the whole work. The efficiency ratio of C and D is \( 4:5 \). Let the efficiencies of C and D be \( 4y \) and \( 5y \) respectively. Thus, their combined efficiency is:
\[ \text{Efficiency of C and D} = 4y + 5y = 9y \]
The time taken by C and D to complete the whole work is:
\[ \text{Time taken by C and D} = \frac{1}{9y} \]

### Step 6: Work Done by C and D in Alternative Days
C and D work alternately starting from C and can complete the total work in \( 44 \frac{1}{2} \) days:
\[ 44 \frac{1}{2} = \frac{89}{2} \text{ days} \]
In \( 44 \frac{1}{2} \) days, they complete the work, which means:
\[ \text{Total Work} = \text{Total Days} \times \text{Combined Efficiency} = \frac{89}{2} \times 9y = \frac{801y}{2} \]
Setting this equal to 1 (total work):
\[ \frac{801y}{2} = 1 \implies y = \frac{2}{801} \]

### Step 7: Substitute Back to Find x
Now we can substitute \( y \) back into the equations. The efficiency of C and D becomes:
\[ \text{Efficiency of C} = 4y = \frac{8}{801}, \quad \text{Efficiency of D} = 5y = \frac{10}{801} \]
Now, we can find \( t \) using the equation from Step 4:
\[ \frac{1}{2x} \times t = \frac{9x - 200}{9x} \]
Since \( t = \frac{1}{9y} \):
\[ \frac{1}{2x} \times \frac{1}{9y} = \frac{9x - 200}{9x} \]
Substituting \( y \):
\[ \frac{1}{2x} \times \frac{801}{18} = \frac{9x - 200}{9x} \]
Cross-multiplying and solving for \( x \):
\[ \frac{801}{36x} = \frac{9x - 200}{9x} \]
Solving this will lead to:
\[ x = \frac{600}{18} = 33.33 \text{ days} \text{ (or } 33 \frac{1}{3} \text{ days)} \]

### Final Answer
Thus, the unique time \( x \) for A to complete the work is:
\[ \boxed{33 \frac{1}{3}} \text{ days} \]