Question
The combined average number of pens with persons βBβ
and βCβ is 75% of the combined average number of pens with persons βAβ and βDβ. The overall average number of pens with all four persons is 45.5, and person βAβ has 16 fewer pens than person βDβ. Based on this information, what is the average number of pens with persons βBβ, βCβ, and βDβ together?Solution
Let the number of pens with βAβ and βDβ be βxβ and βyβ, respectively According to the question, Average number of pens with βBβ and βCβ = {(x + y)/2} Γ 0.75 So, sum of number of pens with βBβ and βCβ = {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 Γ 16 Γ· 7 = 104 Or, x = y β 16 So, y + y β 16 = 104 y = (104 +16)/2 = 120/2 = 60 So, number of pens with βDβ = 60 Number of pens with βAβ = 60 β 16 = 44 So, sum of number of pens with βBβ and βCβ = (44 + 60) Γ 0.75 = 78 So, average number of pens with βBβ, βCβ and βDβ = {(78 + 60)/3} = 138/3 = 46Β
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