ATQ, Let the number of students be 'S' and efficiency of each student be 'e' units/day. Total assignment = S × 12 × e = '12Se' units ATQ, 12Se = S × e + (S - 20 × 1) × e + (S - 20 × 2) × e + (S - 20 × 3) × e + .... + (S - 20 × 15) × e Or, 12S = 16S - 20 × (1 + 2 + 3 + .... + 15) Or, 4S = 20 × 15 × (16/2) [since, sum of first 'n' natural numbers = (n/2) × (n + 1) ] So, 'S' = 600 Let the required time taken by 120 students be 't' days. So, 12 × 600 × e = 120 × t × e Or, 't' = (7,200/120) = 60
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