To determine how many pairs of numbers satisfy the given conditions, we use the relationship between the HCF (highest common factor) and the sum of the numbers.

Given:

1. $\text{HCF} = 30$
2. $a + b = 210$

Solution Steps:

Let the two numbers be $a$ and $b$. Since their HCF is $30$, both numbers must be multiples of $30$. Let:

$a = 30m \quad \text{and} \quad b = 30n$

where $m$ and $n$ are integers such that $\text{gcd}(m, n) = 1$, meaning $m$ and $n$ are coprime.

The sum of the numbers is:

$a + b = 30m + 30n = 30(m + n)$

Given $a + b = 210$:

$30(m + n) = 210 \quad \implies \quad m + n = 7$

Finding Coprime Pairs:

We now find all pairs of integers $(m, n)$ such that $m + n = 7$ and $\text{gcd}(m, n) = 1$. The possible pairs are:

1. $(m, n) = (1, 6)$
2. $(m, n) = (6, 1)$
3. $(m, n) = (2, 5)$
4. $(m, n) = (5, 2)$

Corresponding Pairs of Numbers:

For each pair $(m, n)$, the numbers $a$ and $b$ are:

1. $(m, n) = (1, 6): a = 30 \times 1 = 30, b = 30 \times 6 = 180$
2. $(m, n) = (6, 1): a = 30 \times 6 = 180, b = 30 \times 1 = 30$
3. $(m, n) = (2, 5): a = 30 \times 2 = 60, b = 30 \times 5 = 150$
4. $(m, n) = (5, 2): a = 30 \times 5 = 150, b = 30 \times 2 = 60$

Conclusion:

There are 4 such pairs: $(30, 180), (180, 30), (60, 150), (150, 60)$.

Would you like to see a detailed explanation of why $m$ and $n$ must be coprime?

Questions to Explore Further:

1. Why must $m$ and $n$ be coprime when the HCF of $a$ and $b$ is given?
2. How does the relationship $a + b = 210$ influence the solution process?
3. Can we generalize this method for other sums and HCF values?
4. How would the solution change if $a + b$ were a prime number?
5. What if the condition $\text{HCF}(a, b) = 30$ were replaced with $\text{LCM}(a, b) = 210$?

Tip:

When dealing with HCF problems involving sums, always express the numbers as multiples of the HCF to simplify calculations.