Question

the sum of a 2 digit number is 9 . if the difference between the number formed by reverssing the positions of its digits 27 find the number

Answer 1

Let the digits be a and b
Given , a+b=9 ———[1]

The initial number is = ax10 +b
After reversing the position = bx10 + a


As it is given that the difference between them is 27 ,
Thus,

(10a + b ) - (10b + a ) = 27
10a + b - 10b - a = 27
9a - 9b = 27
9( a - b ) = 27
a - b = 3 ———[2]


Adding equation [1] and [2] ,

a + b = 9
a - b = 3
—————
2a = 12
Therefore,
a = 6

Substituting value of a in [1],
6 + b = 9
Or b = 3

As we know the original number is 10a + b ,
Therefore, the number is
= 10a + b
= 10 x 6 + 3
= 60 + 3
= 63

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

↝ The sum of the digits of a two-digit number is 9.

So,

Let assume that

↝ Digit at one's place be x

and

↝ Digit at tens place is 9 - x.

Tʜᴜs,

Number formed = 10(9 - x) + x × 1 = 90 - 10x + x = 90 - 9x

Reverse number = 1 × (9 - x) + 10x = 9 - x + 10x = 9 + 9x

According to statement,

↝ The difference between the number formed by reverssing the positions of its digits and the original number is 27.

Tʜᴜs,

$\rm :\longmapsto\:9 + 9x - ( 90 - 9x) = 27$

$\rm :\longmapsto\:9 + 9x - 90 + 9x = 27$

$\rm :\longmapsto\:18x - 81 = 27$

$\rm :\longmapsto\:18x = 27 + 81$

$\rm :\longmapsto\:18x = 108$

$\bf\implies \:\boxed{ \tt{ \: x \: = \: 6 \: \: }}$

So,

• Number formed = 90 - 9 × 6 = 90 - 54 = 36