the middle digit of a no between 100 and 1000 is 0 the sum of the other digit is 11. if the digit be reversed the no so formed exeed the original no by 475 find the no..

the digits are htu

t=0
h+u=11
100h+10t+u + 475 = 100u + 10t + h
since t=0,
100h+u + 475 = 100u+h

h+u=11
99h - 99u + 475 = 0

902-209=693 no
803-308=495 no <---?
704-407=297 no
605-506=99 no

Sure there's no typo here?