If mean and median of the distribution $a, b, c, d, e, f$ ( where $b > e > f > c > d > a$) are $20$ and $14$ respectively then mean of numbers $a-f, b-c, e, d$ will be

The given numbers are a, b, c, d, e, f with the constraint b > e > f > c > d > a, and their mean is 20 while their median is 14. The median of six ordered values is the average of the 3rd and 4th terms, meaning (c + d) / 2 = 14, so c + d = 28. The mean is 20, so a + b + c + d + e + f = 120, and substituting c + d = 28 gives a + b + e + f = 92. The required mean of the new set (a - f, b - c, e, d) equals [(a + b + d + e) - (c + f)] / 4, which evaluates to (92 - c - f - c - f) / 4 = (92 - 28 - 2f) / 4 = 16 - 0.5f. Since the exact value of f is not provided, this mean cannot be uniquely determined from the given information, making the correct choice None.

• Thanks
• Wrong explanation