Question

The sum of the digits of a two digit number is 11.If the digits are reversed the number increases by 27.Find the number

Answer 1

let the tenth place be x and the units place be y
the no.=10x+y
              x+y=11   -(1)
if the digits are reversed unit=x and tenth place =y
the no.=10y+x
(10x+y) +27=10y+x
10y+x - 10x-y=27
9y-9x=27
9(y-x)=27
y-x=3(27divided by 3)
-x +y=3  -(2)

x + y = 11
-x+y  =  3
---------------
0+2y=14
 2y=14
y=7
y=7 in equation (1)

x+y=11
x+7=11
x=11-7
x=4

hence the no.=10x+y
                    =10(4)+7
                    =40 + 7
                    =47

Answer 2

let the unit digit be x
since the sum of the digits is 11 , 
the tenth digit is 11 - x
therefore , the number is = (11-x)*10+x
= 110 - 10x + x
= 110 - 9x
if 27 is added , then the number is reversed , 
110 - 9x + 27= 9x + 11
- 9x - 9x = 11 - 110 - 27
- 18x = - 126
x = 126/18
= 7
therefore , the original number is = 110 - 9x 
= 110 - (9*7)
= 110 - 63
= 47