C
Since X stands on V’s shoulder (Both are child acrobats) and M (adult acrobat) and W (child acrobat) stand side by side. Therefore they are on the second level. Hence the possible arrangement is VMW for the second level.
E
As per the conditions in the above statement, if V is fourth, W could be second or third and vice versa. If W is fourth then V would be second or fourth. Hence Y cannot possibly stand on fourth because of either V or W. Also being a child acrobat, he cannot stand first. Hence Y could possibly stand only on the second or third level.
D
A is possible because as per the conditions W and X could be on the third level. B is also possible as M, N and P could be on the first C is also possible as N and V could be on the second level. D is not possible as the condition in the statement. E is also possible if M, Y and Z stand on the second level.
C
The following sets of acrobats could be left standing - N O P Q V and W as a result of M’s fall. Because Q will necessarily be on the second level and in the center he cannot fall nor can N fall since he will necessarily be on the first level and not next to M. Thus by the process of elimination sets A, B, D and E cannot possibly stand that leaves only the set N O P Q V and W standing.
D
If N and Y are standing on M’s shoulder, it presupposes that N and Y are on the second level. Also if Z is on the shoulders of P and O Z too is on the second level. Therefore N Z Y will be on the same level and N and Z must be standing side by side on the same level as one another.
