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