The combined average number of pens with persons ‘B’ and ‘C’ is 75% of the combined average number of pens with persons ‘A’ and ‘D’. The overall average number of pens with all four persons is 45.5, and person ‘A’ has 16 fewer pens than person ‘D’. Based on this information, what is the average number of pens with persons ‘B’, ‘C’, and ‘D’ together?

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