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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

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Video Explanation

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Explanatory Answer

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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The answer is, "4"