In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four
questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at
least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The
remaining 20 marks can be of the following possible combinations. (Four questions of 5
marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15
marks)or (two questions of five marks and one question of ten marks). Hence, the total
number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number
of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is
the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET
has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three
10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten
marks and two 15 marks). This implies, each faculty has given four questions in MT and ET
together. Since it is given that faculty A has given only one question in MT and each of the
other faculties has given more than one question, each of the faculties B, C, D, E and F has
given two questions in MT. This implies faculty A has given three questions in ET and all
other faculties have given two questions each in ET. From the given data we get the
following.

Except A, every other faculty gave at least two questions for MT and all the questions of a
faculty appeared consecutively. Hence, 2nd question in MT is given by F, 4th by C, 10th by D. B
also has given two questions and both appeared consecutively. Hence, 6th and 7th questions
are given by B and the 9th question is given by E. In each test first all 5 marks questions
appeared followed by 10 marks questions and then 15 marks questions. It can be
understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT
questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks
each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2nd question of ET is given by D, the 4th question by C,
6th and 7th questions by A, 9th by E, 10th and 11th by B and 12th by F. Since ET has eight 5
marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of
ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus,
we get the following: