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

Answer

To solve the problem of finding the two numbers whose HCF is 7, LCM is 140, and which lie between 20 and 45, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the relationship between HCF, LCM, and the numbers**:
We know that for two numbers \( A \) and \( B \):
\[ \text{HCF} \times \text{LCM} = A \times B \]
Given that HCF = 7 and LCM = 140, we can substitute these values into the formula:
\[ 7 \times 140 = A \times B \]
This simplifies to:
\[ A \times B = 980 \]

2. **Expressing the numbers in terms of HCF**:
Since the HCF is 7, we can express the two numbers as:
\[ A = 7m \quad \text{and} \quad B = 7n \]
where \( m \) and \( n \) are co-prime integers (i.e., their HCF is 1).

3. **Substituting into the product equation**:
Substituting \( A \) and \( B \) into the product equation gives:
\[ (7m) \times (7n) = 980 \]
Simplifying this, we get:
\[ 49mn = 980 \]
Dividing both sides by 49:
\[ mn = \frac{980}{49} = 20 \]

4. **Finding pairs of co-prime integers**:
We need to find pairs of co-prime integers \( (m, n) \) such that their product is 20. The pairs of factors of 20 are:
- (1, 20)
- (2, 10)
- (4, 5)

Among these pairs, the co-prime pairs are:
- (4, 5)

5. **Calculating the actual numbers**:
Using the co-prime pair \( (4, 5) \):
\[ A = 7m = 7 \times 4 = 28 \]
\[ B = 7n = 7 \times 5 = 35 \]

6. **Checking the range**:
Both numbers 28 and 35 lie between 20 and 45.

7. **Finding the sum of the numbers**:
Now, we can find the sum of the two numbers:
\[ A + B = 28 + 35 = 63 \]

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

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