This is a question based on rules of divisibility. The rule of divisibility of 3 is that the sum of all the digits of the number should be divisible by 3. We have 4 + 2 + 5 + 2 + 7 + 4 + 6 + B = 30 + B, which is completely divisible by three. So B can take value as 0, 3, 6, or 9.

Hence 4 values are possible for B.

Directions for the following 3 (three) items:

Read the following information and answer the three items that follow:

Six students A, B, C, D, E and F appeared in several tests. Either C or F scores the highest. Whenever C scores the highest, then E scores the least. Whenever F scores the highest, B scores the least.

In all the tests they got different marks; D scores higher than A, but they are close competitors; A scores higher than B; C scores higher than A.