The average marks of 6 students in a test is 64 . All the students got different marks, one of the students obtained 70 marks and all other students scored 40 or above. The maximum possible difference between the second highest and the second lowest marks is
Explanation:
Total marks of 6 students = 6 × 64 = 384
Marks of all the students are distinct and greater than 40.
We need to maximise the difference of 2nd highest and 2nd lowest marks
For this we need to minimise the 2nd lowest marks and maximise the 2nd highest marks.
Now, least possible marks of bottom 2 students = 40 and 41. Marks of the students is 70. Let the marks of remaining three students be A, B and C
∴ 40 + 41 + 70 + A + B + C = 384 ⇒ A + B + C = 233.
Now, if C is higest and B is second highest, we can maximise B by minimising A. Least possible marks of A now is 42. ⇒ B + C = 191
For B to be maximum possible, B and C should be as close as possible. ∴ C = B + 1
⇒ B + B + 1 = 191 ⇒ B = 95
∴ The 6 numbers are: 40, 41, 42, 70, 95, 96
⇒ Differnce between 2nd highest and 2nd lowest marks = 95 - 41 = 54.
Hence, option (b).
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