The HCF of 64, 48 and y is 8. Which of the options below cannot be a possible value of y? 

Correct option is B

Given

Numbers: 64, 48, and y

HCF of 64, 48, and y = 8

Concept Used

The HCF of three numbers is the greatest number that divides all three numbers.

If the HCF is 8, then y must be divisible by 8 and share no higher common factor with 64 and 48 other than 8.

Solution

Calculating the HCF of 64, 48, and each option y:

Option (A). y = 88: HCF(64, 48, 88) = 8 → Valid.

Option (B). y = 96: HCF(64, 48, 96) = 16 → Not Valid (HCF exceeds 8).

Option (C). y = 104: HCF(64, 48, 104) = 8 → Valid.

Option (D). y = 72: HCF(64, 48, 72) = 8 → Valid.

Thus, The option that cannot be a possible value of y is Option(B): 96.