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The maximum marks were 100 and no student obtained full marks. A’s marks were just higher than that of B, whose marks are perfect cube. Only two students got higher marks than B. With the first hint it is clear that even the person securing highest marks did not get 100 marks. With respect to second and third hints it is clear that B received the third highest marks and A received the second highest marks. The possible marks for B are 8, 27 and 64. The difference between the marks of C and B is 22. C’s marks are in even number. With the given hints it is clear that B’s marks are 64, only then second hint would get satisfied. C’s marks could be 22 less than that of B or more than. 
F received higher marks than only D, who received 23 marks. E’s marks are half of the marks secured by C. With respect to the first hint it is clear that D received the lowest and F received the second lowest marks. E’s marks were less than that of C and we know that A and B received second and third highest marks respectively. __ > A > B > E > F > D Thus the only left person C must have got the highest marks. With this we can say that Case-1 fails. 
F’s marks were 7 more than the difference between the marks secured by E and D. The sum of the marks secured by all the students was 320. As per first hint, the marks secured by F are 27. At present the sum of marks obtained by all the students except A is 243. So the marks obtained by A is 320 – 243 = 77. 
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