Given: In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47.
Asked: For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by
Let the scores in increasing order be x1, x2, .... x10 where x1 is lowest score and x10 is highest score.
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 42*9 = 378
x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 47*9 = 423
x10 - x1 = 423 - 378 = 45
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)max is when x1 = 42; and x10 = 42 + 45 = 87
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)max = 378 + 87 = 465
Mean_max = 465/10 = 46.5
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)min is when x10 = 47
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)min = 378 + 47 = 425
Mean_min = 425/10 = 42.5
Difference = Mean_max - Mean_min = 46.5 - 42.5 = 4
IMO C
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com