If 10 men and 15 women complete a piece of work in 8 days while 12 men and 8 women can complete the same piece of work in 10 days. If A boy who is 50% less efficient than the man, can do the same work in 50 days.
Quantity I: Time taken by 2 men, 4 women and 18 boys to complete the work.
Quantity II: Time taken by 9 men, 3 women and 6 boys to complete the same work.

Options

A. Quantity I ` gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `ge `Quantity II
D. Quantity I = Quantity II or No relation

Answer

To solve the problem step by step, we will first determine the efficiency of men, women, and boys based on the information provided.

### Step 1: Determine the work done by men and women

We know:
- 10 men and 15 women complete the work in 8 days.
- 12 men and 8 women complete the work in 10 days.

Let the work done by 1 man in 1 day be \( m \) and the work done by 1 woman in 1 day be \( w \).

From the first scenario:
\[ 10m + 15w = \frac{1}{8} \quad \text{(since they complete the work in 8 days)} \]
This can be rewritten as:
\[ 80m + 120w = 1 \quad \text{(multiplying through by 8)} \]

From the second scenario:
\[ 12m + 8w = \frac{1}{10} \quad \text{(since they complete the work in 10 days)} \]
This can be rewritten as:
\[ 120m + 80w = 1 \quad \text{(multiplying through by 10)} \]

### Step 2: Set up the equations

Now we have two equations:
1. \( 80m + 120w = 1 \) (Equation 1)
2. \( 120m + 80w = 1 \) (Equation 2)

### Step 3: Solve the equations

We can solve these equations simultaneously. Let's multiply Equation 1 by 3 and Equation 2 by 2 to eliminate \( w \):

From Equation 1:
\[ 240m + 360w = 3 \quad \text{(multiplying by 3)} \]

From Equation 2:
\[ 240m + 160w = 2 \quad \text{(multiplying by 2)} \]

Now subtract the second from the first:
\[ (240m + 360w) - (240m + 160w) = 3 - 2 \]
\[ 200w = 1 \implies w = \frac{1}{200} \]

Substituting \( w \) back into Equation 1 to find \( m \):
\[ 80m + 120 \left(\frac{1}{200}\right) = 1 \]
\[ 80m + \frac{120}{200} = 1 \]
\[ 80m + \frac{3}{5} = 1 \]
\[ 80m = 1 - \frac{3}{5} = \frac{2}{5} \]
\[ m = \frac{2}{400} = \frac{1}{200} \]

### Step 4: Determine the efficiency of boys

A boy is 50% less efficient than a man, so:
\[ b = \frac{m}{2} = \frac{1/200}{2} = \frac{1}{400} \]

### Step 5: Calculate Quantity I

For Quantity I, we need to find the time taken by 2 men, 4 women, and 18 boys:
\[ \text{Total work done per day} = 2m + 4w + 18b \]
Substituting the values:
\[ = 2 \left(\frac{1}{200}\right) + 4 \left(\frac{1}{200}\right) + 18 \left(\frac{1}{400}\right) \]
\[ = \frac{2}{200} + \frac{4}{200} + \frac{18}{400} \]
\[ = \frac{6}{200} + \frac{18}{400} \]
Converting \(\frac{18}{400}\) to a common denominator:
\[ \frac{18}{400} = \frac{9}{200} \]
So,
\[ = \frac{6}{200} + \frac{9}{200} = \frac{15}{200} = \frac{3}{40} \]

The time taken to complete the work:
\[ \text{Time} = \frac{1}{\text{Total work done per day}} = \frac{1}{\frac{3}{40}} = \frac{40}{3} \text{ days} \]

### Step 6: Calculate Quantity II

For Quantity II, we need to find the time taken by 9 men, 3 women, and 6 boys:
\[ \text{Total work done per day} = 9m + 3w + 6b \]
Substituting the values:
\[ = 9 \left(\frac{1}{200}\right) + 3 \left(\frac{1}{200}\right) + 6 \left(\frac{1}{400}\right) \]
\[ = \frac{9}{200} + \frac{3}{200} + \frac{6}{400} \]
Converting \(\frac{6}{400}\) to a common denominator:
\[ \frac{6}{400} = \frac{3}{200} \]
So,
\[ = \frac{9}{200} + \frac{3}{200} + \frac{3}{200} = \frac{15}{200} = \frac{3}{40} \]

The time taken to complete the work:
\[ \text{Time} = \frac{1}{\text{Total work done per day}} = \frac{1}{\frac{3}{40}} = \frac{40}{3} \text{ days} \]

### Conclusion

Both quantities are equal:
\[ \text{Quantity I} = \text{Quantity II} = \frac{40}{3} \text{ days} \]

Thus, the answer is:
**Quantity I = Quantity II**