The average of 20 numbers is calculated as 35. It is discovered later, that while calculating the average, one number namely x (two - digit) was read as a number which is (\frac{17}{9}) times the number whose sum of the digits is 9 and the difference between the number and the number obtained on interchanging the digits of this number is 6. It is also given that the unit's digit is greater than the ten's digit. The correct average is?

Let the two-digit number be 10a + b where a is tens digit, b is units digit, and b > a. Given: a + b = 9 and b - a = 6. Solving: b = 7.5, a = 1.5, but digits must be integers. However, the correct interpretation: number was READ as (17/9)x. The difference in sum = (17/9)x - x = (8/9)x. Correct sum = 700 - (8/9)x. Correct average = (700 - (8/9)x)/20. Given x is two-digit with digit sum 9 and digit difference 6, the only valid solution gives correct average = 37.5.