Question

Sum of the digits of a two-digit number is 11 . The given number is less then the number obtained by interchanging the digits by 9 . Find the number .​

Answer 1

Answer:

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Step-by-step explanation:

Let the number be 

Then 

According to the question

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

10x + y = 10y + x-

10x-x + y-10y = -9

9x-9y = -9

x-y = -1⋯(ii)

Adding both equations (i) and (ii)

x + y + x-y = 11-1

2x = 10

x = 5

Put value of x in equation (i)

x + y = 11

5 + y = 11

y = 11-5

y = 6

Hence the number is xy = 56

Answer 2

Let the digit at ones place be y and at the tens place be x

Given: x + y = 11 ........(1)

Original number= 10x + y

Reversed number= 10y + x

Given: 10x+y=(10y+x)-9

10x-x+y-10y=-9

9x-9y=-9

9 (x-y)=-9

x-y=-9/9

x-y=-1 ........(2)

Adding (1) and (2),

x+y+x-y=10

x=5

Substituting in (1),

y=6

Thus the number is 56.