Question

a number consists of two digit whose sum is 8 . if 18 is added to the number its digits is reversed .find it

Answer 1

Hi ,

Let ten's place digit = x ,

Units place digit = 8 - x ,

The number = 10x + 8 - x

= 9x + 8 -----( 1 )

Reversing the digits the new number

is so formed = 10( 8 - x ) + x

= 80 - 10x + x

= 80 - 9x----( 2 )

According to the problem given ,

( 1 ) + 18 = ( 2 )

9x + 8 + 18 = 80 - 9x

9x + 26 = 80 - 9x

9x + 9x = 80 - 26

18x = 54

x = 54/18

x = 3

Therefore ,

Put x = 3 in equation ( 1 ) , we get

The number = 9x + 8

= ( 9 × 3 ) + 8

= 27 + 8

= 35

I hope this helps you.

:)

Answer 2

Let x be the 1st digit number and y be the 2nd digit number.

Therefore the required 2-digit number = 10x + y.  ------ (*)

Given that their sum is 8.

x + y = 8  ---- (1)

Given that if 8 is added to the number its digits is reversed.

10x + y + 18 = 10y + x

9x - 9y = - 18

x - y = -2.   ------ (2)


On solving (1) & (2), we get

x + y = 8

x - y = -2

---------------

2x = 6

x = 3.


Substitute x = 3 in (1), we get

x + y = 8

3 + y = 8

y = 8 - 3

y = 5.

Substitute in (*), we get

The required number = 10(x) + y

                                     = 10(3) + 5

                                     = 30 + 5

                                     = 35.


Verification:

3 + 5 = 8

8 = 8.


10x + y  +18 = 10y + x

10(3) + 5 + 18 = 10(5) + 3

30 + 5 + 18 = 50 + 3

53 = 53.


Hope this helps!