40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child-7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again?

The ring starts with Child-1. The passing pattern is: Child-1 passes to Child-2 (jump of 1) Child-2 passes to Child-4 (jump of 2) Child-4 passes to Child-7 (jump of 3) And so on. The jump increases by 1 each time.

Let 'k' be the number of changes (passes). The total displacement (sum of jumps) after 'k' changes is S_k = 1 + 2 + 3 + ... + k. Using the formula for the sum of the first 'k' natural numbers, S_k = k * (k + 1) / 2.

For the ring to be back in the hands of Child-1, the total displacement must be a multiple of the total number of children (40). So, S_k must be a multiple of 40. k * (k + 1) / 2 = 40 * m (where 'm' is an integer) k * (k + 1) = 80 * m

We need to find the smallest positive integer 'k' such that k * (k + 1) is a multiple of 80. Let's test the values: If k = 14: k * (k + 1) = 14 * 15 = 210. 210 is not a multiple of 80. If k = 15: k * (k + 1) = 15 * 16 = 240. 240 is a multiple of 80 (240 = 3 * 80).

Thus, after 15 changes, the ring will be back in the hands of Child-1.

The final answer is B) 15.