Discussion

Question:

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 2^nd highest and 2^nd lowest marks

For this we need to minimise the 2^nd lowest marks and maximise the 2^nd 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 2^nd highest and 2^nd lowest marks = 95 - 41 = 54.

Hence, option (b).

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