Instructions

Answer the following questions 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.

Question 32

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

Solution

We can see that Bhama got calls from all the college hence it is essential for her to clear all 4 sectionals along with overall cutoff. That essentially means that she has to score marks which are greater or equal to the greatest cut-off for that section across the six colleges. from the table we can see that to clear section A the cut offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. Hence, in order to clear the sectional cut-off of section A for all the colleges, she should have scored at least 45 marks.

Similarly, the minimum marks required sectional cut-off for section B for all colleges = Max{42, 42, 41} = 45.

Similarly, the minimum marks required sectional cut-off for section C for all colleges = Max{42, 45, 46, 43} = 46.

Similarly, the minimum marks required sectional cut-off for section D for all colleges = Max{45, 44} = 45.

So total minimum aggregate marks required is 45+45+46+45 = 181.

Note we have to check for aggregate cut-off as well for all 6 colleges. We can see that 181 is also more than aggregate cut-off marks of all six colleges. Therefore option B is the correct answer.

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