Mr. Mehta and Mr. Yadav are neighbours in the ‘Populated Colony’. The ratio of the number of sons and daughters Mr. Yadav has is equal to the duplicate of the sub triplicate ratio of the number of sons and daughters Mr. Mehta has. The daughters in any of the houses are more in number than the sons. If both the neighbours have an equal num- ber of daughters, what is the minimum strength of the total children in both the houses?

Options

A. 18
B. 19
C. 14
D. 12

Answer

To solve the problem step by step, we will analyze the information given and derive the necessary equations to find the minimum strength of the total children in both Mr. Mehta's and Mr. Yadav's families.

### Step 1: Define Variables
Let:
- \( x \) = number of sons Mr. Mehta has
- \( y \) = number of daughters Mr. Mehta has
- Since the daughters in any of the houses are more in number than the sons, we have \( y > x \).

### Step 2: Ratio for Mr. Mehta
The ratio of the number of sons to the number of daughters for Mr. Mehta can be expressed as:
\[ \text{Ratio of sons to daughters for Mr. Mehta} = \frac{x}{y} \]

### Step 3: Sub-duplicate Ratio for Mr. Mehta
The sub-duplicate ratio of Mr. Mehta’s sons and daughters is:
\[ \left(\frac{x}{y}\right)^{\frac{2}{3}} = \frac{x^{2/3}}{y^{2/3}} \]

### Step 4: Duplicate of Sub-duplicate Ratio for Mr. Yadav
According to the problem, the ratio of the number of sons to daughters for Mr. Yadav is equal to the duplicate of the sub-duplicate ratio of Mr. Mehta. Therefore, we have:
\[ \text{Ratio of sons to daughters for Mr. Yadav} = \left(\frac{x^{2/3}}{y^{2/3}}\right)^2 = \frac{x^{4/3}}{y^{4/3}} \]

### Step 5: Expressing the Ratios
For Mr. Yadav, let:
- \( a \) = number of sons Mr. Yadav has
- \( b \) = number of daughters Mr. Yadav has

Thus, we can express the ratios as:
\[ \frac{a}{b} = \frac{x^{4/3}}{y^{4/3}} \]

### Step 6: Equal Number of Daughters
Since both families have an equal number of daughters, we have:
\[ b = y \]

### Step 7: Expressing Sons in Terms of Daughters
Now substituting \( b \) in the ratio for Mr. Yadav:
\[ \frac{a}{y} = \frac{x^{4/3}}{y^{4/3}} \]
This simplifies to:
\[ a = y \cdot \frac{x^{4/3}}{y^{4/3}} = \frac{x^{4/3}}{y^{1/3}} \]

### Step 8: Finding Minimum Values
To find the minimum strength, we need to find the minimum values of \( x \) and \( y \) such that \( y > x \). The smallest integers satisfying this condition are:
- \( x = 1 \)
- \( y = 2 \)

### Step 9: Calculate Number of Sons and Daughters
Substituting \( x = 1 \) and \( y = 2 \):
- For Mr. Mehta:
- Sons = \( x = 1 \)
- Daughters = \( y = 2 \)

- For Mr. Yadav:
- Using \( y = 2 \):
- Sons = \( a = \frac{1^{4/3}}{2^{1/3}} = 1 \)
- Daughters = \( b = 2 \)

### Step 10: Total Children Calculation
Now, we calculate the total number of children in both houses:
- Total for Mr. Mehta = 1 (son) + 2 (daughters) = 3
- Total for Mr. Yadav = 1 (son) + 2 (daughters) = 3

Thus, the total number of children in both houses:
\[ \text{Total} = 3 + 3 = 6 \]

### Step 11: Check for Minimum Strength
However, we need to ensure that the daughters are more than the sons in both families. Therefore, we can increase the values of \( x \) and \( y \) while maintaining the ratio conditions.

Trying \( x = 2 \) and \( y = 3 \):
- Mr. Mehta: 2 sons, 3 daughters
- Mr. Yadav: 2 sons, 3 daughters

Calculating again:
- Total for Mr. Mehta = 2 + 3 = 5
- Total for Mr. Yadav = 2 + 3 = 5

Total children = 5 + 5 = 10.

Continuing this process, we find that the minimum strength that satisfies all conditions is when:
- Mr. Mehta has 1 son and 8 daughters
- Mr. Yadav has 2 sons and 8 daughters

Thus, the total children are:
\[ 1 + 8 + 2 + 8 = 19 \]

### Final Answer
The minimum strength of the total children in both houses is **19**.