Discussion

Explanation:

The aggregate cut-off for each college is given in the common data. In order for Charlie to get minimum marks in one of the sections, he should have got maximum marks (i.e. 50) in the other three sections.

For example, the aggregate cut-off in college 1 is 176. Since, we want minimum marks in a section he should have gotten an aggregate of exactly 176. To minimise one of the sections, assume that he got 50 marks in the 3 sections whose cut-off is given in the common data. Then, Charlie will get a call from college 1 if he gets at least 176 – (50 × 3) = 26 marks in section D, provided that the cut-off for this section is also 26.

Now, there is at least one unknown sectional cut-off for each of the colleges, so we can use the same logic as used above for each of the remaining colleges.

For college 2, the minimum marks that Charlie needs to get a call = 175 – 150 = 25

For college 3, the minimum marks that Charlie needs to get a call = 171 – 150 = 21

For college 4, the minimum marks that Charlie needs to get a call = 178 – 150 = 28

For college 5, the minimum marks that Charlie needs to get a call = 180 – 150 = 30

For college 6, the minimum marks that Charlie needs to get a call = 176 – 150 = 26

The question states that Charlie only gets a call from 2 of the colleges. So, Charlie got 25 marks.

Hence, option (c).

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