If the HCF and LCM of two consecutive (positive) even numbers be 2 and 84 respectively, then the sum of the numbers is:

Options

A. 30
B. 26
C. 14
D. 34

Answer

To solve the problem, we need to find two consecutive positive even numbers whose HCF (Highest Common Factor) is 2 and LCM (Least Common Multiple) is 84. Let's denote the two consecutive even numbers as \(2p\) and \(2q\).

### Step-by-Step Solution:

1. **Understanding HCF and LCM**:
- The HCF of \(2p\) and \(2q\) is given as 2. This implies that the HCF of \(p\) and \(q\) must be 1 (since we can factor out 2).
- The LCM of \(2p\) and \(2q\) is given as 84.

2. **Using the relationship between HCF and LCM**:
- The relationship between HCF and LCM for two numbers \(a\) and \(b\) is given by:
\[ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \]
- Here, substituting \(a = 2p\) and \(b = 2q\):
\[ 2 \times 84 = (2p) \times (2q) \]
- Simplifying this gives:
\[ 168 = 4pq \implies pq = \frac{168}{4} = 42 \]

3. **Finding pairs of \(p\) and \(q\)**:
- We know that \(p\) and \(q\) are consecutive integers and their product is 42. Therefore, we need to find pairs of integers that satisfy this condition.
- The pairs of factors of 42 are:
- \(1 \times 42\)
- \(2 \times 21\)
- \(3 \times 14\)
- \(6 \times 7\)

4. **Identifying consecutive integers**:
- Among the pairs listed, \(6\) and \(7\) are consecutive integers. Thus, we can take \(p = 6\) and \(q = 7\).

5. **Calculating the even numbers**:
- Now substituting back to find the even numbers:
\[ 2p = 2 \times 6 = 12 \]
\[ 2q = 2 \times 7 = 14 \]

6. **Finding the sum of the numbers**:
- Finally, we calculate the sum of the two even numbers:
\[ 12 + 14 = 26 \]

### Conclusion:
The sum of the two consecutive positive even numbers is **26**.