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A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

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 Hint : If the number is ab it can also be written as 10a+b since a is at tens place and b is at ones place.

Let the digits at the tens place be x and ones place be y.
Hence the number is 10x + y ……(i)
By reversing the number it becomes 10y + x ……(ii)
According to the question we have
x + y = 5 ……(iii)
Also when 9 is added to the number the digits get interchanged.
$
  10x + y + 9 = 10y + x \\
  9x - 9y + 9 = 0\, \\
 $
$y - x = 1\,\,\,$ ……(iv)
Adding (iii) &(iv) we get,
$
  2y = 6 \\
  y = 3 \\
    \\
$ (v)
Then x can be calculated from (iii)or(iv) we get,
x = 2 (vi)
From (i),(v),(vi) the number is
10(2) + 3 = 23
Hence the number is 23.

Note :- In these types of problems of finding numbers we have to assume variables and assign them a place it may be ones, tens, hundredth etc. Then according to the question make the equation and solve it to get the number.