Question

A number is consist of two digits whose sum is five when the digits are reversed the number become greater by nine . find the number ​

Answer 1

Let the digits in the tens place be x and ones place be y

Hence the number is (10x + y)

Number formed by reversing the digits = (10y + x)

Given sum of digits = 5

⇒ x + y = 5 → (1)

Also given that when 9 is added to the number the digits gets interchanged

(10x + y) + 9 = (10y + x)

⇒ 10x + y + 9 – 10 y – x = 0

⇒ 9x – 9y = –9

⇒ x – y = –1 → (2)

Add (1) and (2), we get

x + y = 5

x – y = –1

————

2x = 4

∴ x = 2

Put x = 2 in x + y = 5

2 + y = 5

∴ y = 3

Hence the number is 23

Answer 2

Answer:

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

Hence the number is 10x+y

By reversing 10y+x

Sum of digit =5

⇒x+y=5⟶(i)

Also that when 9 is added to the number the digits get interchanged.

∴(10x+y)+9=(10y+x)

10x+y+9=10y+x=0

9x−9y=−9

x−y=−1⟶(ii)

Adding (i) &(ii) we get ,

⇒x+y=5

⇒x−y=−1

⇒2x=4

⇒x=2

Put x=2 in x+y=5

∴2+y=5

⇒y=3

Hence the no: is 23.