Let the 5 children be named M, N, P, Q and R. assign minimum possible values starting from M up to P. Q and N will then have the highest possible values.

M = 1 trophy
N = 2 trophies
P = 3 trophies

At this point M+N+P = 1+2+3 = 6. Therefore Q +R will be 30-6 = 24. Since Q and R cannot have the same number of trophies: Q will be 11 trophies and R 13 trophies.

Q = 11 trophies (second-highest number of trophies )
R = 13 trophies

Hence option B (11 trophies) is the answer.

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We can let the three children who have the least number of trophies have 1, 2, and 3 trophies, respectively, since no child can have the same number of trophies as any other sibling. Thus, these three children have a total of 1 + 2 + 3 = 6 trophies.

So we are left with a total of 30 - 6 = 24 trophies for the two children with the greatest number of trophies. Since we want the second-most number of trophies to be a large as possible, we will ensure that the greatest number of trophies is as small as possible while still retaining its role as the greatest number. We see that a 12-12 split almost works, but the two numbers are not allowed to be the same value.

Thus, we can let the child with the greatest number of trophies have 13 trophies, and hence, the child with the second most trophies has 11.

Answer: B
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