If marks of F is 80 and sum of marks of F and G is 130,

Solution

F + G = 120 80 + G = 130 G = 130 – 80 = 50. Possible Marks of I will between 50 and 80. The marks of E is more than the marks of H. The number of persons more than E is one less than the number of persons less than H. E > ___ >  _____ > ___ H > ___ ……… (I) ____ > E > ___ >  H > ___ > ___ ……… (II) The marks of F is more than I, whose marks is more than G. At least two persons have less marks than G. E > F >  I > G > H > D ……… (I) ____ > E > ___ >  H > ___ > ___ ……… (II) Case 2 will get discarded. Therefore, the final arrangement is as follow: E(highest) > F > I > G > H > D(lowest)