Question

The sum of the digits of a two digit number is 15, if the number formed by reversing the digits is less than the original number by 27, Find the original number ?

Answer 1

heya folk

let tens digit of the original number be x

so original number,

10(x) + (15-x) 

reversing digits mean

10(15-x) + x

therefore

[10(x) + (15-x)]-[10(15-x)+x] = 27

 10x + 15-x - 150+10x-x = 27

10x + 10x + 15 - 150 -x - x = 27

20x - 135 -2x = 27

18x - 135 = 27

18x = 27 + 135

18x = 162

x = 162/18

x = 9

original number = 10(x) + (15-x) 

                              = 10(9) + (15-9)

                              = 90+6

                              = 96

Answer 2

Let the digits in the units place be x and the digits in tens place be y .

Therefore, The number is in the form 10y + x

Given,

x + y = 15

=> x = 15 - y...................(1)


Also, The number formed by reversing the digits is less than the original number by 27

i.e. 10x + y = (10y + x) - 27

=> 10x + y = 10y + x - 27

=> 10x - x + y - 10y +27 = 0

=> 9x - 9y + 27 = 0

=> 9 ( x - y + 3 ) = 0

=> x - y + 3 = 0

=> 15 - y - y + 3 = 0 [From (1)]

=> -2y = -3 - 15

=> y = 9

Putting the value of y = 9 in eq.(1) we get

=> x = 15 - 9

=> x = 6

Thus, the original number is 10(9) + 6 i.e. 96