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Let the number of pens with ‘A’ and ‘D’ be ‘x’ and ‘y’, respectively According to the question, Average number of pens with ‘B’ and ‘C’ = {(x + y)/2} × 0.75 So, sum of number of pens with ‘B’ and ‘C’ = {3(x + y)/4} So, sum of number of pens with all 4 persons = (x + y) + {3(x + y)/4} = (7x + 7y)/4 According to the question, (7x + 7y)/16 = 45.5 Or, (x + y) = 45.5 × 16 ÷ 7 = 104 Or, x = y – 16 So, y + y – 16 = 104 y = (104 +16)/2 = 120/2 = 60 So, number of pens with ‘D’ = 60 Number of pens with ‘A’ = 60 – 16 = 44 So, sum of number of pens with ‘B’ and ‘C’ = (44 + 60) × 0.75 = 78 So, average number of pens with ‘B’, ‘C’ and ‘D’ = {(78 + 60)/3} = 138/3 = 46