Question

the sum of the number and the number obtained by reversing the digit is 121 if one digit is 3 more than the other find the number

Answer 1

Let the unit digit be x
and ten's digit be y


Original number :- 10y + x

Reversed number :- 10x + y


according to question ,

10y + x + 10x + y = 121

11y + 11x = 121

x + y = 11

x = 11 - y ........ ( i )


* If one digit is three more than other!!

then we have to consider larger and smaller number!!

Let x be larger
and y be smaller


x = y + 3 ...... ( ii )


Equating both ( i ) and ( ii )


11 - y = y + 3

11 - 3 = y + y

8 = 2y

4 = y



Now,

putting the value of y in ( ii )

x = y + 3

x = 4 + 3

x = 7



The required number is

10y + x

10× 4 + 7

= 40 + 7

= 47


So, the required number is 47!!