University Admission Cut-offs

CAT 2008 Slot 1 · DILR · Medium · Data Interpretation

All CAT 2008 Slot 1 questionsMore Data Interpretation questionsAll CAT DILR

Passage / data set

For admission to various affiliated colleges, a university conducts a written test with four different sections, each with a maximum of 50 marks. The following table gives the aggregate as well as the sectional cut-off marks fixed by six different colleges affiliated to the university.

CollegeSection ASection BSection CSection DAggregate Cut-off
College 1424242-176
College 2-4545-175
College 3--46-171
College 443--45178
College 545-43-180
College 6-41-44176

Question 1 of 3

Bhama got calls from all colleges. What could be the minimum aggregate marks obtained by her?

  1. A.

    180

  2. B.

    181

  3. C.

    196

  4. D.

    176

  5. E.

    184

Answer

B

Explanation

To clear all sectional cut-offs, Bhama must score at least:

  • Section A: max⁡(42,43,45)=45\max(42, 43, 45) = 45
  • Section B: max⁡(42,45,41)=45\max(42, 45, 41) = 45
  • Section C: max⁡(42,45,46,43)=46\max(42, 45, 46, 43) = 46
  • Section D: max⁡(45,44)=45\max(45, 44) = 45

Sum of minimum sectional marks =45+45+46+45=181= 45 + 45 + 46 + 45 = 181. This also satisfies all aggregate cut-offs (max aggregate cut-off is 180). Thus, minimum aggregate marks =181= 181.

Question 2 of 3

Charlie got calls from two colleges. What could be the minimum marks obtained by him in a section?

  1. A.

    0

  2. B.

    21

  3. C.

    25

  4. D.

    35

  5. E.

    41

Answer

C

Explanation

To minimize marks in one section, maximize marks in the other 3 sections to 50 each. Lowest aggregate required for 2 calls is 175 (Colleges 2 and 3). 175−(50+50+50)=25175 - (50 + 50 + 50) = 25. With scores 50, 50, 50, 25 in Sections A, B, C, D respectively, Charlie clears College 2 and College 3 cut-offs.

Question 3 of 3

Aditya did not get a call from even a single college. What could be the maximum aggregate marks obtained by him?

  1. A.

    181

  2. B.

    176

  3. C.

    184

  4. D.

    196

  5. E.

    190

Answer

C

Explanation

To maximize marks while getting zero calls, fail at least one sectional requirement for every college with minimum deduction. Score 50 in Section A, 50 in Section B, 41 in Section C (fails College 1, 2, 3, 5 cut-offs), and 43 in Section D (fails College 4, 6 cut-offs). Total aggregate =50+50+41+43=184= 50 + 50 + 41 + 43 = 184.

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace