Question

The sum of digits of a two digit number is 11. The number obtained by interchanging the digits of the given number exceeds that number by 63. Find the number

Answer 1

Let the two digits of a no.be x and y

Difference between them is 11

So x+y=11

x+y–11=0........(1)

x=11–y

A two digit no. Will have a no. In ten's place and a no. At one's place

So the no. Is 10x+y

Now according to the question

On interchanging the digits

Means 10x+y becomes 10y+x

And on getting interchanged the new no. Exceeds old no. By 63

So 10x+y+63=10y+x

10x–x+y–10y+63=0

9x–9y+63=0

9(x–y+7)=0

x–y+7=0.....(2)

Substituting the value of x=11–y in equation (2)

We get

11–y–y+7=0

11+7=2y

18=2y

y=18/2

y=9

Substituting the value of y in equation 1

We get

x+y–11=0

x+9–11=0

x=11–9

x=2

Substituting the values of x and y in 10x+y

We get 10×2+9=29

Interchanged no. Will be 10y+x

We get 10×9+2=92

The new no. Exceeds old no. By 63 as 92–29=63.

And sum of digits is 11

As 9+2=11

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Answer 2

Step-by-step explanation:

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