Question

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

Answer 1

 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.
$$\begin{gathered} 10x+y+9=10y+x \\ 9x-9y+9=0 \end{gathered}$$
$y-x=1$ ……(iv)
Adding (iii) &(iv) we get,
$$\begin{gathered} 2y=6 \\ y=3 \end{gathered}$$ (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.