Question

27. 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. I want whole calculation

Answer 1

Answer:

let the number is XY.

Then X+Y=11................... (i)

A.T.Q

10X+Y=(10Y+X)-9

10X+Y=10Y+X-9

10X-X+Y-10Y=-9

9X-9Y=-9

9 (X-Y)=-9

X-Y=-9/9

X-Y=-1 .............................. (ii)

adding both equation, we get..

X+Y=11

- ( X-Y)=-1

_________

2Y=12

Y=12/2

Y=6

Put the value of Y in equation (i)

X+Y=11

X+6=11

X=11-6

X=5

let the number is XY.

Then X+Y=11................... (i)

A.T.Q

10X+Y=(10Y+X)-9

10X+Y=10Y+X-9

10X-X+Y-10Y=-9

9X-9Y=-9

9 (X-Y)=-9

X-Y=-9/9

X-Y=-1 .............................. (ii)

adding both equation, we get..

X+Y=11

- ( X-Y)=-1

_________

2Y=12

Y=12/2

Y=6

Put the value of Y in equation (i)

X+Y=11

X+6=11

X=11-6

X=5

Answer 2

Let the digit at ones place be x

it is given that sum of the digit is 11

tens digit= (11-x)

Thus,

The original number =10(11-x)+x

The original number=110-9x

• On interchanging the digit of the given number the digit at omes place become (11-x) and didgit at tens place become x

New Number= 10x+(11-x)=9x+11.

It is given that The number is less then the number obtained by interchanging the digits by 9

So,

New Number-(original number)=9

9x+11-(110-9x)=9

9x+11-110+9x=9

18x-99=9

18x=108

x=$\frac{\cancel{108}}{\cancel{18}}$

x=6

Now put the value of Number

Ones digit= 6

tens digit=5

So, The number is 56.