Question

A student needs to enroll for a minimum of 60 credits. A student cannot enroll for more than 70 credits. The credits are divided amongst project and three distinct sets of courses namely, core courses, specialization courses, and elective courses. It is compulsory for a student to enroll for exactly 15 credits of core courses and exactly 20 credits of project. In addition, a student has to enroll for a minimum of 10 credits of specialization courses. The maximum credits of elective courses that a student can enroll for is ______

• A. 10
• B. 15
• C. 20
• D. 25

Hint:

Maximize total credits and minimize specialization credits to leave the most room for electives.

Answer 1

The Correct Option is D

Step 1: List the fixed and variable credit components. Let Core = 15 credits (fixed, compulsory), Project = 20 credits (fixed, compulsory), Specialization = S credits (variable, with S >= 10), and Elective = E credits (variable, the quantity to maximize). The total enrolled credits T = Core + Project + S + E = 15 + 20 + S + E = 35 + S + E.

Step 2: Write down the given constraints. The total credits must satisfy 60 <= T <= 70 (minimum 60, maximum 70). The specialization credits must satisfy S >= 10 (minimum 10, with no stated upper cap on specialization alone). We need to find the maximum possible value of E.

Step 3: Express E in terms of T and S. From T = 35 + S + E, we get E = T - 35 - S.

Step 4: Maximize E by choosing T and S optimally. To make E as large as possible, we want T as large as possible and S as small as possible, since E decreases as either T decreases or S increases. The largest allowed T is 70 (the upper credit cap), and the smallest allowed S is 10 (the specialization minimum). Substituting these extreme values: E_max = 70 - 35 - 10 = 25.

Step 5: Verify the resulting allocation is feasible. With Core = 15, Project = 20, Specialization = 10, Elective = 25, the total is 15 + 20 + 10 + 25 = 70, which lies exactly at the upper bound of the allowed range 60 <= T <= 70, so it is a valid enrollment. The specialization value of 10 also meets its minimum requirement of S >= 10. All constraints are satisfied simultaneously, confirming that E = 25 is achievable, not just an unreachable upper bound.

Step 6: Rule out the other options. Option (A) 10, option (B) 15, and option (C) 20 are all smaller than 25 and therefore not the maximum - the student could enroll for more elective credits than these values while still staying within the rules, so they understate the true maximum.

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