The average marks of 6 students in a test are 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

Total marks = 6 * 64 = 384. One student = 70. Remaining 5 students sum = 314. To maximize the difference between 2nd highest (H2) and 2nd lowest (L2), we minimize L2 and maximize H2. Let L1=40, L2=40. Let H1=x, H2=y. Remaining marks = 384 - 70 - 40 - 40 = 234. To maximize y, we minimize x. Since x > y, let x = y + 1. 2y + 1 = 234. y = 116.5. This is impossible as marks are usually <= 100. Assuming max marks 100: H1=100, H2=94. Remaining = 384 - 70 - 100 - 94 = 120. L1=40, L2=40. Difference = 94 - 40 = 54.