Discussion

Explanation:

Since HCF of first two numbers is 48, let the numbers be 48a and 48b.

Also, HCF of next two numbers is 36, let the numbers be 36x and 36y.

Now, HCF of all 4 numbers = HCF(48a, 48b, 36x, 36y).

Since, a and b are co-primes and also x and y are also co primes, a, b, x and y will have not common factor.

⇒ HCF(48a, 48b, 36x, 36y) = HCF(48, 36) = 12.

Note: HCF of all the four numbers = HCF of (HCF of the first 2 numbers and HCF of the other 2 numbers.

Hence, option (c).

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