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In the previous question, what can be the minimum and maximum value of A’s average if 5 teams are transferred from B to A?
20.8 and 22.4
20.6 and 22.6
20.8 and 22.8
20.8 and 21.8
For minimum, teams with minimum score would be transferred from B to A So, total score transferred = 24 * 5 = 120 A(Minimum Average) = (20 x 20 + 120) /25 = 20.8 For maximum, if we take teams with maximum score, total score transferred = 32 * 5 = 160 However, if we assume that 5 teams scored the highest score, the average of remaining teams in B would come out to be, (625 - 160) /20 = 465/20 = 23.25 which is lower than the lowest score of B. So, 5 teams in B can’t have score of 32. so, minimum total score which need to be on group B = 24 * 20 = 480 Score that can be transferred = 625 – 480 = 145 A(Maximum Average) = (400 + 145)/25 = 545/25 = 21.8
By: Amit Kumar ProfileResourcesReport error
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