HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45 the sum of the numbers is

Options

A. 70
B. 77
C. 63
D. 56

Answer

To solve the problem, we need to find two numbers whose HCF (Highest Common Factor) is 7 and LCM (Lowest Common Multiple) is 140, and which lie between 20 and 45.

### Step-by-Step Solution:

1. **Understanding HCF and LCM**:
- Given HCF = 7 and LCM = 140.
- The relationship between HCF, LCM, and the two numbers (let's call them \( a \) and \( b \)) is given by:
\[ HCF \times LCM = a \times b \]
- Therefore, we can write:
\[ 7 \times 140 = a \times b \]
- This simplifies to:
\[ a \times b = 980 \]

2. **Expressing the Numbers**:
- Since the HCF is 7, we can express the two numbers as:
\[ a = 7x \quad \text{and} \quad b = 7y \]
- Here, \( x \) and \( y \) are coprime (i.e., their HCF is 1).

3. **Substituting into the Product**:
- Substituting \( a \) and \( b \) into the product equation:
\[ (7x) \times (7y) = 980 \]
- This simplifies to:
\[ 49xy = 980 \]
- Dividing both sides by 49 gives:
\[ xy = \frac{980}{49} = 20 \]

4. **Finding Pairs of \( (x, y) \)**:
- We need to find pairs of integers \( (x, y) \) such that \( xy = 20 \) and \( x \) and \( y \) are coprime.
- The pairs of factors of 20 are:
- \( (1, 20) \)
- \( (2, 10) \)
- \( (4, 5) \)

5. **Checking Coprimeness**:
- Among these pairs, we check for coprimeness:
- \( (1, 20) \): Coprime
- \( (2, 10) \): Not coprime
- \( (4, 5) \): Coprime
- Valid pairs are \( (1, 20) \) and \( (4, 5) \).

6. **Calculating the Numbers**:
- For \( (x, y) = (1, 20) \):
\[ a = 7 \times 1 = 7, \quad b = 7 \times 20 = 140 \quad \text{(not in range)} \]
- For \( (x, y) = (4, 5) \):
\[ a = 7 \times 4 = 28, \quad b = 7 \times 5 = 35 \quad \text{(both in range)} \]

7. **Finding the Sum**:
- The two numbers are 28 and 35.
- The sum of the numbers is:
\[ 28 + 35 = 63 \]

### Final Answer:
The sum of the two numbers is **63**.