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Question 3 : 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. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by
Here a2 to a9 is common to both the terms
So, a1 + (a2 to a9) = 42 × 9
a10 + (a2 to a9) = 47 × 9
Solving these two a10 - a1 = 45
a1, a2, a3,………………………, a9, (a1 + 45)
One instance is every number is 42
42, 42, …………………, 42, 42+45 (a1 to a9 are equal) -----------(1)
Another instance is every number is 47
47 – 45, 47, …………………, 47, 47 (a2 to a10 are equal) ---------(2)
Mean of (1) = 46.5
Mean of (2) = 42.5
(1) – (2) = 4
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