Question

the sum of the digits of a two digit number is 4. the number got by interchanging the digits is 18 less than the original number. what is the number?​

Answer 1

Solution :-

Let once digit be x

Let ten's digit be y

Here,

The sum of the two digit number = 4

x + y = 4 ...........eq( 1 )

Therefore,

Required number = 10x + y

Now, Digit are reversed

New no = 10y + x

According to the question,

( 10x + y) - ( 10y + x) = 18

10x + y - 10y - x = 18

9x - 9y = 18

9( x - y) = 18

x - y = 18/9

x - y = 2 .............eq( 2 )

Adding eq( 1.) and eq( 2 )

( x + y) + ( x - y) = 4 + 2

2x = 6

x = 6/2

x = 3

Now, Subsitute the value of x in eq( 1 )

x + y = 4

3 + y = 4

y = 4 - 3 = 1

Hence, The two digit number be

31 $.$

Answer 2

Solution :-

Let once digit be x

Let ten's digit be y

Here,

The sum of the two digit number = 4

x + y = 4 ...........eq( 1 )

Therefore,

Required number = 10x + y

Now, Digit are reversed

New no = 10y + x

According to the question,

( 10x + y) - ( 10y + x) = 18

10x + y - 10y - x = 18

9x - 9y = 18

9( x - y) = 18

x - y = 18/9

x - y = 2 .............eq( 2 )

Adding eq( 1.) and eq( 2 )

( x + y) + ( x - y) = 4 + 2

2x = 6

x = 6/2

x = 3

Now, Subsitute the value of x in eq( 1 )

x + y = 4

3 + y = 4

y = 4 - 3 = 1

Hence, The two digit number be

31 ..