Hi NandishSS,

Best way to solve these types of questions is to list down few possibilities then build upon with the given information.

Given: Product of digits of a three digit number is 120.

Generally any digit place can take up a integer value from 0 to 9(Except "0" cannot be in the first place of any digit number).

But in this question none of the digit places can take zero (because that will make the product as zero). Also since it's a multiple of 5, then one of the digits must be multiple of 5.

Let k = 5xy

5 * x * y = 120.

So x* y = 24, so possibilities could be 6 and 4 or 8 and 3.

So possibilities of the number could be 853 or 546 etc..

So we got a sense now, for this to be divisible by 5. We should know in certain that unit's place has to be 5.

Let's check the statements,

Statement I is insufficient:

K is a multiple of 3.

RULE: If the sum of all the digits of a number is multiple of 3, then the number is multiple or divisible by 3.

Here, we can't take 8, 5 and 3 combinations.

But even if we take the combinations of 5, 4 and 6 , still we are not sure whether the units place of k is 5 or not. It could be 654 or 465

So insufficient.

Statement II is sufficient:

To think of, we can write it from least to greatest from left to right,
853 or 654 etc.. In any case 5 cannot be in the units place. Otherwise it will contradict the given statement(because 5 is not the least digit in k).

So we can certainly say that k is not divisible by 5.

So sufficient.

So the answer is B.

Hope this helps.