Direction: Answer the question based on the following information.For...
For University of BerkleyMinimum aggregate = 176
Minimum marks in Section A = 45
Minimum marks in Section B = 42
Minimum marks in Section C = 42
Minimum marks in Section D can be obtained if Maximum marks are obtained in Sections A, B and C
Therefore Minimum marks in section D can be = 176 – 150 = 26
For University of Massachusetts
Minimum Aggregate = 175
Minimum marks in Section B = 46
Minimum marks in Section C = 46
Minimum marks in section A or D can be obtained if maximum are obtained in the other three.
Therefore Minimum marks in section A or D = 175 – 150 = 25
For University of Michigan
Minimum Aggregate = 180
Minimum marks in section D = 48
Minimum Marks can be obtained in Section A, B or C if Maximum are obtained in Section D
Therefore Minimum marks in Section A, B or C = 180 – 150 = 30
For University of Minnesota
Minimum Aggregate = 181
Minimum marks in Section C = 41
Minimum marks in Section D =44
Minimum marks in section A or B can be obtained if maximum are obtained in the other 3
Therefore minimum marks in Section A or B = 181 – 150 = 31
The minimum marks that can still make a student eligible for selection are 25 in University of Massachusetts.