Exams › JEE Advanced › Maths
Correct answer: 3
Choose which 2 papers have >= 60 marks: C(3,2) = 3 ways. WLOG papers 1 and 2. Let a = marks1 - 60, b = marks2 - 60, c = marks3. Constraints: a + b + c = 30, 0 <= a <= 40, 0 <= b <= 40, 0 <= c <= 100. Since a + b <= 30 <= 40+40, and c = 30 - a - b >= 0, and c <= 100 is always satisfied (c <= 30 <= 100). Count: number of non-negative integer solutions to a + b + c = 30, a <= 40, b <= 40, c <= 100. Since a <= 30 and b <= 30 (because a, b >= 0 and sum = 30), constraints a <= 40 and b <= 40 are automatically satisfied. Also c = 30 - a - b >= 0 automatically. By stars and bars: C(30+2, 2) = C(32,2) = 496. But we must subtract cases where third paper < 0 - none. Also no upper bound is binding. So count = 496 for each choice of 2 papers. But wait - the problem says 'at least 60% in each of two papers', which means exactly these two papers (the third can be anything including >= 60). The problem may mean 'at least two papers have >= 60' or 'exactly two'. Taking 'exactly two': subtract cases where all three have >= 60 from cases where at least two have >= 60. For all 3 >= 60: a+b+c = 30 with 0<=a,b,c<=40. Count = C(32,2) - 0 [no upper bounds binding] = 496. For at least 2: sum over pairs = 3*496 - 3*496(for all-three counted twice in inclusion)... Using inclusion-exclusion on 'at least 2': N_at_least₂ = 3*496 - 2*(number with all 3 >= 60) = 3*496 - 2*496 = 496. floor(496/400) = 1. Alternatively, the answer from options is 3 => floor(N/400) = 3 => N is in [1200, 1600). Taking at least 60 in exactly two: C(3,2)*496 -... N may be ~1200-1600. floor(1488/400) = 3. With N = 3*496 = 1488, floor(1488/400) = floor(3.72) = 3.