Quant - Challenging problem

Hello all,

I wanted to share this question as I liked it very much and this would be a treat for quantitative reasoning lovers!

Question:

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 ___.

I have provided the answer with steps below and it is blurred as it is a spoiler!

The reasoning can be made like this:

Sum of least 9 marks = 42*9 = 378

Sum of greatest 9 marks = 47*9 = 423

=> Range = highest mark - lowest mark = 423 - 378 = 45. (Equation 1)

Hence, an average of the 10 numbers can be written as:

(least mark + sum of greatest 9 numbers)/10 = (least mark+ 423)/10 (Equation 2)

In order to maximize the mean, the minimum value of the highest mark should be 47. Because, if the highest mark is reduced from 47, then the mean of the highest 9 marks can never be 47! (Note that the marks are in ascending order!) Hence, if listed in ascending order, the marks should be like - least mark, 47, 47,… 47 (10 marks). Here, the highest mark is 47.

From equation 1, the least mark = highest mark - 45 = 47 - 45 = 2.

Hence, from equation 2, minimum mean = (2 + 423) / 10 = 42.5!

Similarly, in order to minimize the mean, the maximum value of the least mark should be 42 due to the same logic that the least mark can never be less than 42. If it is less than 42, then the average of the least 9 marks can never be 42! Hence, the marks would appear like - 42, 42, … highest mark (10 marks). Here, the lowest mark is 42.

From equation 1, highest mark = lowest mark + 45 = 42 + 45 = 89.

Hence, from equation 2, maximum mean = (42 + 423)/10 = 46.5!

Finally, the result = maximum mean - minimum mean = 4!!!

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