Question

A number consists of two digits whose sum is 8 . if 18 is added to the number its digits gets reversed find the number

Answer 1

Let,

Unit place digit = y

Tens place digit = x

Number formed = 10x + y

After reversing the digits, we get

= 10y + x

A/q

x + y = 8 ___________( 1 )

Also,

10x + y + 18 = 10y + x

10x - x + y - 10y = - 18

9x - 9y = - 18

9 ( x - y ) = - 18

x - y = - 2 ___________( 2 )

Add ( 1 ) and ( 2 ), we get

( x + y ) + ( x - y ) = 8 + ( - 2 )

2x = 6

x = 3

Put this value in ( 1 ), we get

3 + y = 8

y = 5

Required number -

10 ( 3 ) + 5 = 35

^^"

Answer 2

Let the digit at the units and tens place be x and (8-x) respectively.

The original number= 80-9x

The number with digits reversed= 9x-8

Acc. to question,

80-9x+18=9x-8

98-9x=9x-8

18x=106

x=7

The required number is 72