**Problem Analysis**

We need to find the number of possible values of X, such that the highest common factor (HCF) of 9000 and X is equal to 60.

To solve this problem, we can use the fact that the HCF of two numbers is always a divisor of both numbers. Therefore, the HCF of 9000 and X must be a divisor of 9000.

**Finding the Divisors of 9000**

To find the number of possible values of X, we need to find the divisors of 9000.

The prime factorization of 9000 is:
9000 = 2^3 * 3^2 * 5^3

To find the divisors of 9000, we can use the fact that any divisor of 9000 can be formed by taking a combination of the prime factors 2, 3, and 5.

Therefore, we can choose any power of 2 from 0 to 3, any power of 3 from 0 to 2, and any power of 5 from 0 to 3.

This gives us a total of (3+1) * (2+1) * (3+1) = 4 * 3 * 4 = 48 possible divisors of 9000.

**Counting the Divisors that are Divisible by 60**

Out of these 48 divisors, we need to count the ones that are divisible by 60.

A number is divisible by 60 if it is divisible by both 2 and 3 and ends with a 0 or 5.

Therefore, we need to count the divisors that have a power of 2 greater than or equal to 2, a power of 3 greater than or equal to 1, and a power of 5 greater than or equal to 1.

For the power of 2, we have 3 choices: 2^2, 2^3, 2^0.
For the power of 3, we have 2 choices: 3^1, 3^0.
For the power of 5, we have 2 choices: 5^1, 5^0.

Therefore, the number of divisors that are divisible by 60 is 3 * 2 * 2 = 12.

**Finding the Number of Possible Values of X**

Since the HCF of 9000 and X is 60, and the divisors that are divisible by 60 are the possible values of X, we can conclude that there are 12 possible values of X.

Therefore, the correct answer is (a) 24.

**Summary**

The number of possible values of X, such that the HCF of 9000 and X is equal to 60, is 24. This is because the HCF of two numbers is always a divisor of both numbers, and the divisors of 9000 can be formed by taking a combination of the prime factors 2, 3, and 5. Out of these divisors, we need to count the ones that are divisible by 60, which gives us a total of 24 possible values of X.