Question

a number consists of two digits whose sum is 10 if 18 is added to the number then the digit change their place find the number ​

Answer 1

Answer:

Given:

•  sum of digits  is 10
• 18 is added to the number then the digit change their place find the number ​

Solving Question:

Let the digits of the number be 'x' and 'y'.Sum is 10

⇒ x +y =10 ......equ(1)

Again, 18 is added to the number then the digit change their place find the number (Let the number be 10x+y )

⇒  10x+y + 18 = 10y+x ..........equ(2)

Solution:

 simplify equ(2)

10x +y+18 = 10y+x

⇒ 9x -9y = -18   (multiply -1 )

9y - 9x = 18 ........equ(3)

9(x + y = 10 ) ⇒ 9x +9y = 90 ........equ(4)

equate equ(4) and equ(3)

9y - 9x = -18

9y + 9x = 90

______________

we get, 18y = 72

or, y = 4

substitute the value in equ(1)

x + y = 10

x + 4 = 10

or, x = 6         number =  (10x+y)

Thus the digits are 6 and 4

The number is 64

Answer 2

Let Tens Place digit=x

Let Unit Place digit=y

Sum of Digits=10

x+y=10

x=10-y

Original Number=10x+y

Reverse Number=10y+x

If 18 is added to Original number it become reverse number

10x+y+18=10y+x

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

18=10y+x-10x-y

18=9y-9x

18=9(y-x)

18/9=y-x

2=y-x

Y-X=2

x=10-y

Y-(10-y)=2

y-10+y=2

2y-10=2

2y=2+10

2y=12

y=12/2

y=6

We know

X=10-y

x=10-6

x=4

x=4

y=6

Number=10x+y

=10×4+6

=46

Original Number=46