Question

the sum of a digit of a two digit number is 11 if the number obtained by reversing the digit is 9 less than the original number find the number

Answer 1

let the digit be :(10a+b)

Now. Acc. to que.
a+b=11 →[1] (sum of digits)
and
10b+a= 10a+b - 9 (on reversing the digits)
→10b-b + a-10a + 9 =0
→9b-9a+9=0(dividing throughout by 9)
→b-a+1=0
→b+1=a →[2]

substituting Eq.[2] in Eq[1]
We get:
(b+9) +b =11
→2b+1=11
→2b=11-1
→2b=10
→b=10/2
→b=5 →[3]

Sub. Eq.[3] in Eq.[2]
→5+1=a
→a=6

So the Digits are a=6and b=5

Checking the correctness of the answers:

the sum of 6 and 5=11

and 65-56=9

Answer 2

Answer:

65

Step-by-step explanation:

Given:

Sum of two digit number = 11

Let unit’s digit be ‘x’

and tens digit be ‘y’,

then x + y=11…(i)

and number = x+10y

By reversing the digits,

Unit digit be ‘y’

and tens digit be ‘x’

and number =y+10x+9

Now by equating both numbers,

y+10x+9=x+10y

10x+y–10y–x=−9

9x–9y=−9

x–y=−1…(ii)

Adding (i) and (ii), we get

2x=10

x=10/2

=5

∴y=1+5=6

By substituting the vales of x and y, we get

Number = x+10y

=5+10×6

=5+60

=65

∴ The number is 65.