Let one of the two digits of the number be given as #x#. let this be the number in tens place. Then the number in ones place will be #10-x# as it is given that sum of the two digits is 10.
Now the number can be written as :
#=>10xx( x) + 1xx(10-x)#
#=> 10x +(10-x)# -------------let this be (1)
If we reverse the digits , it becomes :
#10xx(10-x) +1xx x#
#=> 10(10-x)+x# ---------let this be (2)
Given that: (2)-(1) = 54:
#=> [10(10-x)+x] - [10x +(10-x)] = 54#
#=> [ 100 - 10x +x] - [ 10x +10 -x] =54#
#=> [100 -9x] - [9x+10] =54#
#=> 100-9x-9x-10=54#
#=> 90 - 18x =54#
#=> -18x = 54-90#
#=> -18x = -36#
#=> x=2#
So the original two digit number can be given from(1) as:
#10x +(10-x) = 10(2) +(10-2)#
#=> 20 +8 =28#
The number is #28#
Cross check:
reversing digits, we get #82#
and
#82-28 = 54#
