Question

the sum of the digits of a two digit number is 9 .number formed by interchanging the digits is 45 more than the original number .find the original number and check the solution

Answer 1

Heyaa user!!!!

Here's your answer :-

Let the digit in the one's place be y and the digit in the ten's place be x...

Now, acc to question,

x+y=9. .........(i)

Number formed =10x+y.....

Now ,no. formed by interchanging the digits will be 10y+x.....

Acc to question,

10y+x-10x-y=45

=>9y-9x=45

=>y-x=5. .............(ii)

Adding eq i and ii we get :-

x+y+y-x=9+5

=>2y=14

=>y=7

Now, putting value of x in eq i we get :-

x+7=9

=>x=9-7=2......

Required no. = 10x+y....
=27

Answer 2

Answer:

let the tens digit = x

ones digit = y

x+y=9

10x+y......origunal no.

10y+x......new no. or by interchanging

According to question

10y+x=(10x+y)+45

y-x=5

by animation we get x=7

7+y=9

y=2

original no.27