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Question 25 : The average marks of 6 students in a test is 64 . All the students got different marks, one of the students ohtained 70 marks and all other students scored 40 or above. The maximum possible difference between the second highest and the second lowest marks is

- 50
- 54
- 57
- 58

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Let the marks of the students be a < b < c < d < e < f

Need maximum possible difference, So e_{max} - b_{min}

So, a has to be 40, then b = 41

\\frac{sum of all marks}{6}) = 64

Sum of the marks = 384

E should be as high as possible, If c is 70 then d has to be more than 70, So take d = 70, then c = 42

Then a + b + c + d = 40 + 41 + 42 + 70 = 193

Then e + f = 384 - 193 = 191

E < F, E is as high as possible, So divide it closely

Then e = 95 , f = 96

So, e - b = 95 - 41 = 54.

Choice B is the correct answer.

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