A number 'p' is such that it is divisible by 7 but not by 2. Another number 'q' is divisible by 6 but not by 5, then the following experssion which necessarily be an interger is:

Options

A. `(7p+6q)/42`
B. `(5p+6q)/71`
C. `(6p+7q)/42`
D. none of these

Answer

To solve the problem, we need to analyze the conditions given for the numbers \( p \) and \( q \) and then evaluate the expressions provided to determine which one results in an integer.

### Step 1: Understand the conditions for \( p \) and \( q \)

- \( p \) is divisible by 7 but not by 2. This means \( p \) can be expressed as:
\[ p = 7k \quad \text{(where \( k \) is an odd integer)} \]

- \( q \) is divisible by 6 but not by 5. This means \( q \) can be expressed as:
\[ q = 6m \quad \text{(where \( m \) is any integer that does not make \( q \) a multiple of 5)} \]

### Step 2: Evaluate the expressions

We need to evaluate the following expressions to see which one is guaranteed to be an integer:

1. **Expression A:**
\[ \frac{6p + 7q}{42} \]

2. **Expression B:**
\[ \frac{5p + 6q}{35} \]

3. **Expression C:**
\[ \frac{7p + 6q}{42} \]

### Step 3: Substitute \( p \) and \( q \) into the expressions

#### For Expression A:
\[ 6p + 7q = 6(7k) + 7(6m) = 42k + 42m = 42(k + m) \]
Thus,
\[ \frac{6p + 7q}{42} = \frac{42(k + m)}{42} = k + m \]
This is an integer.

#### For Expression B:
\[ 5p + 6q = 5(7k) + 6(6m) = 35k + 36m \]
Thus,
\[ \frac{5p + 6q}{35} = \frac{35k + 36m}{35} = k + \frac{36m}{35} \]
This is not necessarily an integer since \( \frac{36m}{35} \) may not be an integer.

#### For Expression C:
\[ 7p + 6q = 7(7k) + 6(6m) = 49k + 36m \]
Thus,
\[ \frac{7p + 6q}{42} = \frac{49k + 36m}{42} \]
This is not guaranteed to be an integer since \( \frac{49k + 36m}{42} \) may not simplify to an integer.

### Conclusion

The only expression that is guaranteed to be an integer is:
\[ \frac{6p + 7q}{42} \]

### Final Answer
The expression which necessarily be an integer is:
\[ \frac{6p + 7q}{42} \]