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ATQ, Let the efficiency of each boy be 'b' units/day. Total work = x × 18 × b = '18xb' units ATQ: 18xb = x × b + (x - 15 × 1) × b + (x - 15 × 2) × b + (x - 15 × 3) × b + .... + (x - 15 × 23) × b Or, 18x = 24x - 15 × (1 + 2 + 3 + .... + 23) Or, 6x = 15 × (23/2) × (23 + 1) [since, sum of first 'n' natural numbers = (n/2) × (n + 1) ] So, 'x' = 690 Therefore, required value = 0.2 × 690 = 138
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