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.
10 men start another work 'Y' and completed that work in certain days. How many women and children should be assigned to do the same work in same number of days as 10 men take. Give that number of children working is double of the number of women they all work with same individual efficiency on work 'Y' as they work on original work.

Options

A. 4 women, 8 child
B. 5 women, 10 child
C. 15 women, 30 child
D. 10 women, 20 child

Answer

To solve the problem step by step, we will first determine the efficiencies of men, women, and children based on the information given in the question.

### Step 1: Determine the total work done by men and women
We know that:
- 10 men and 12 women can complete the work in 5 days.

Let the work done by one man in one day be \( m \) and the work done by one woman in one day be \( w \). The total work can be expressed as:
\[ \text{Total Work} = \text{Number of Workers} \times \text{Days} = (10m + 12w) \times 5 \]
Thus, the total work \( W \) is:
\[ W = 50m + 60w \quad \text{(1)} \]

### Step 2: Determine the total work done by women and children
We also know that:
- 2 women and 6 children can complete the same work in 32 days.

Let the work done by one child in one day be \( c \). The total work can also be expressed as:
\[ W = (2w + 6c) \times 32 \]
Thus, the total work \( W \) is:
\[ W = 64w + 192c \quad \text{(2)} \]

### Step 3: Equate the two expressions for total work
From equations (1) and (2), we can equate them:
\[ 50m + 60w = 64w + 192c \]
Rearranging gives:
\[ 50m = 4w + 192c \quad \text{(3)} \]

### Step 4: Work done by men, women, and children
We know from the problem statement that:
- 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.

This gives us:
\[ 6m = 8w + 8c \]
Dividing through by 2:
\[ 3m = 4w + 4c \quad \text{(4)} \]

### Step 5: Solve the system of equations
Now we have two equations (3) and (4):
1. \( 50m = 4w + 192c \)
2. \( 3m = 4w + 4c \)

From equation (4), we can express \( m \) in terms of \( w \) and \( c \):
\[ m = \frac{4w + 4c}{3} \]

### Step 6: Substitute \( m \) into equation (3)
Substituting this expression for \( m \) 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:
\[ 188w = 376c \]
Thus, simplifying gives:
\[ \frac{w}{c} = \frac{2}{1} \quad \text{(5)} \]

### Step 7: Determine the ratio of efficiencies
From equation (5), we can conclude that:
- The ratio of the efficiencies of women to children is \( 2:1 \).

### Step 8: Calculate the number of women and children for work 'Y'
Let \( y \) be the number of women assigned to work 'Y'. Then the number of children assigned will be \( 2y \).

The efficiency of \( y \) women and \( 2y \) children is:
\[ \text{Efficiency} = yw + 2yc = yw + 2y \left(\frac{w}{2}\right) = yw + yw = 2yw \]

The efficiency of 10 men is:
\[ \text{Efficiency of 10 men} = 10m = 10 \left(\frac{4w + 4c}{3}\right) = \frac{40w + 40c}{3} \]

### Step 9: Equate the efficiencies
Since both groups complete the work in the same number of days, we can set their efficiencies equal:
\[ 2yw = 10m \]
Substituting \( m \):
\[ 2yw = 10 \left(\frac{4w + 4c}{3}\right) \]
This simplifies to:
\[ 2yw = \frac{40w + 40c}{3} \]
Cross-multiplying gives:
\[ 6yw = 40w + 40c \]
Substituting \( c = \frac{w}{2} \) from the ratio:
\[ 6yw = 40w + 20w \]
Thus:
\[ 6yw = 60w \implies y = 10 \]

### Step 10: Calculate the number of children
Since the number of children is double the number of women:
\[ \text{Number of children} = 2y = 20 \]

### Final Answer
Thus, the required number of women and children to complete the work in the same number of days as 10 men is:
- **10 women and 20 children.**