10 men and 12 women can complete a work in 5 days and 2 women and 6 children can complete the same work in 32 days. The work done by 6 men in one day is equal to the work done by 8 women and 8 children together in one day.
If 8 men and 4 women start another work 'X' and they worked for 5 days and completed 1/3 rd of the work. In how many days the remaining work is done by 8 women 4 children. They all work with same individual efficiency on work 'X' as they work on original work

Options

A. 25 days
B. 20 days
C. 23 days
D. 10 days

Answer

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Calculate the total work done in terms of men, women, and children.

We know:
- 10 men and 12 women can complete the work in 5 days.
- 2 women and 6 children can complete the same work in 32 days.

Let the work done by 1 man in one day be \(M\), by 1 woman be \(W\), and by 1 child be \(C\).

From the first condition:
\[ \text{Total Work} = (10M + 12W) \times 5 \]
\[ \text{Total Work} = 50M + 60W \quad \text{(1)} \]

From the second condition:
\[ \text{Total Work} = (2W + 6C) \times 32 \]
\[ \text{Total Work} = 64W + 192C \quad \text{(2)} \]

Equating (1) and (2):
\[ 50M + 60W = 64W + 192C \]
Rearranging gives:
\[ 50M = 4W + 192C \quad \text{(3)} \]

### Step 2: Establish the relationship between men, women, and children.

From the problem, we know:
The work done by 6 men in one day is equal to the work done by 8 women and 8 children together in one day:
\[ 6M = 8W + 8C \]
Dividing through by 2:
\[ 3M = 4W + 4C \quad \text{(4)} \]

### Step 3: Solve equations (3) and (4).

From equation (4), we can express \(M\) in terms of \(W\) and \(C\):
\[ M = \frac{4W + 4C}{3} \]

Substituting \(M\) into equation (3):
\[ 50\left(\frac{4W + 4C}{3}\right) = 4W + 192C \]
Multiplying through by 3 to eliminate the fraction:
\[ 200W + 200C = 12W + 576C \]
Rearranging gives:
\[ 200W - 12W = 576C - 200C \]
\[ 188W = 376C \]
\[ \frac{W}{C} = \frac{376}{188} = 2 \]
Thus, \(W = 2C\).

### Step 4: Substitute back to find the relationship with \(M\).

Substituting \(W = 2C\) back into equation (4):
\[ 3M = 4(2C) + 4C \]
\[ 3M = 8C + 4C = 12C \]
\[ M = 4C \]

### Step 5: Find the total work in terms of \(C\).

Substituting \(M\) and \(W\) into equation (1):
\[ 50(4C) + 60(2C) = \text{Total Work} \]
\[ 200C + 120C = \text{Total Work} \]
\[ \text{Total Work} = 320C \]

### Step 6: Calculate the work done by 8 men and 4 women in 5 days.

The work done by 8 men and 4 women in one day:
\[ \text{Work in one day} = 8M + 4W = 8(4C) + 4(2C) = 32C + 8C = 40C \]
In 5 days:
\[ \text{Total work done} = 5 \times 40C = 200C \]

### Step 7: Calculate the remaining work.

The total work is \(320C\), and they completed \(200C\):
\[ \text{Remaining work} = 320C - 200C = 120C \]

### Step 8: Calculate how many days it takes for 8 women and 4 children to complete the remaining work.

The work done by 8 women and 4 children in one day:
\[ \text{Work in one day} = 8W + 4C = 8(2C) + 4C = 16C + 4C = 20C \]

Now, to find the number of days \(d\) to complete the remaining work:
\[ 20C \times d = 120C \]
\[ d = \frac{120C}{20C} = 6 \text{ days} \]

### Final Answer:
The remaining work will be completed by 8 women and 4 children in **6 days**.