There are 6 persons in a list A, B, C, D, E and F. They got different marks in an examination. C got more marks than D but less than B. E got more marks than F and F got the least marks among them. The one who got the highest marks got 400 marks and the one who got the least marks got 60 marks. The one who got the highest marks, his position in the list is number 1 and the one who got the least marks, his position in the list is number 6.
If A got 400 marks and E’s position in the list is `2^(nd)` last and the persons who got last 3 positions in the list, their average marks is 95, then what will be the total marks of D and E together?

Options

A. 245
B. 235
C. 225
D. 200

Answer

C

Explanation

From this information their positions are in decreasing order-
`a (400) gt B gt C gt D gt E gt F (60)`
It is given D, E and F have average marks = 95 `xx 3 = 285`
F got 60 marks so-
D+E=285-60= 225 marks