Question

The sum of a two digit number and the number obtain by reversing the order if its digits is 121 and the two digits differ by 3. Find the numbers.

Answer 1

let  digits are be x and y  

number  obtained = 10y +x  

number obtained to reversing the digits = 10x+y  

sum =  121

10y+x +10x+y = 121

11x +11y = 121

11[x+y ] = 121

x+y = 11  (i)

x -- y = 3(i)

y = 7  ;;;     x = 4     then number = 47

Answer 2

Step-by-step explanation:

✞︎ Given :-

The sum of two digit number and the number obtained by reversing the order of its digits is 121

And the two digits differ by 3

❦︎ To find :-

The numbers........

✯︎ Solution :-

Let the two numbers be 'x' and 'y'

➪︎ Then the number is 10x + y

The number obtained by reversing the digits is yx

➪︎ Then the number is 10y + x

❁︎ According to problem,

Sum of them is 121....

➪︎ So, 10x + y + 10y + x = 121

➪︎ 11x + 11y = 121

➪︎ 11 ( x + y ) = 121

➪︎ x + y = 11.......(1)

⍟︎ And also given that,

The numbers are differ by 3

➪︎ So, x - y = 3.........(2)

By adding (1) & (2)

➪︎ x + y = 11

➪︎ x - y = 3

___________

2x = 14

x = 7

Substitute x = 7 in (1) ,

➪︎ 7 + y = 11

➪︎ y = 11 - 7

➪︎ y = 4

So, the numbers is xy = 74 ; yx = 47

✰︎ so the required number is 74 or 47