CAT 2008 — DILR Question 2
Answer the following question based on the information given below.
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. A student will get admission only if he/she gets marks greater than or equal to the cut-off marks in each of the sections and his/her aggregate marks are at least equal to the aggregate cut-off marks as specified by the college.
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Bhama got calls from all colleges. What could be the minimum aggregate marks obtained by her?
Answer & solution
- A
180
181
- C
196
- D
176
- E
184
Since Bhama got calls from all colleges, she must have cleared each of the 4 sections. This means that for a particular section she scored more marks than the greatest cut-off for that section across the six colleges.
For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The greatest cut-off among them is 45.
So, in order to clear the sectional cut-off of section A for all the colleges, she should have scored at least 45.
Since we wish to minimise her marks, we should take her score in section A as 45.
Similarly, in sections B, C and D, she scored 45, 46, and 45 marks respectively.
∴ Bhama’s minimum marks such that she gets calls from all the colleges = 45 + 45 + 46 + 45 = 181
Hence, option (b).
Note: This is already greater than the highest aggregate cut-off of all colleges (which is 180 for college 5). So, she will get calls from all 6 colleges.