If the HCF (GCD) of two positive integers is 8, then which of the following values can NEVER be their LCM? (Choose the one option that is impossible.)

Introduction / Context:
This question tests the relationship between HCF (GCD) and LCM for two numbers. If two numbers have HCF = 8, both numbers must be multiples of 8. A key fact is: for numbers x and y, x * y = HCF(x,y) * LCM(x,y). Also, if we write the numbers in reduced form, the LCM becomes tightly constrained.

Given Data / Assumptions:

• HCF of the two numbers is 8
• We must decide which candidate LCM value is impossible

Concept / Approach:
Let the two numbers be 8a and 8b, where gcd(a,b) = 1 (after factoring out the HCF, the remaining parts are coprime). Then:
LCM(8a, 8b) = 8 * a * b
Because a and b are integers, the LCM must be a multiple of 8. A necessary condition is that any LCM value must be divisible by 8.

Step-by-Step Solution:
Step 1: If HCF = 8, write the numbers as 8a and 8b with gcd(a,b) = 1.Step 2: LCM becomes 8ab, so the LCM must be divisible by 8.Step 3: Check each option for divisibility by 8:32 ÷ 8 = 4 (divisible)48 ÷ 8 = 6 (divisible)60 ÷ 8 is not an integer (not divisible)152 ÷ 8 = 19 (divisible)120 ÷ 8 = 15 (divisible)Step 4: The only value that is not a multiple of 8 is 60, so it can never be the LCM.

Verification / Alternative check:
Construct examples: (8, 32) gives HCF 8 and LCM 32; (8, 48) gives LCM 48; (8, 152) gives LCM 152; (8, 120) gives LCM 120. No pair of numbers with HCF 8 can have LCM 60, because 60 is not a multiple of 8.

Why Other Options Are Wrong:
32: Possible with numbers such as 8 and 32.48: Possible with numbers such as 8 and 48.152: Possible with numbers such as 8 and 152.120: Possible with numbers such as 8 and 120.

Common Pitfalls:
Forgetting that both numbers must be multiples of the HCF.Not checking divisibility of the candidate LCM by the HCF.Confusing the roles of HCF and LCM in the product formula.

Final Answer:
60