Question

a number consists ot two digits the sum of the digits being 9. if 9 is added to the number the digits are reversed. find the number

Answer 1

Let the unit digit be x and tens digit be 9-x
No formed =(9-x)*10+x
90-10x+x=90-9x
Reversed no= 10x+(9-x)
9x+9
According to the question,
90-9x+9=9x+9
90+9-9=9x+9x
90=18x
90/18=x
5=x
Number=(9-5)*10+5
4*10+5
45

Answer 2

let,
the number be 10x+y
the reversed number be 10y+ x
where x and y are the digits

case 1: Sum of digits is 9
y + x = 9 ----->1

case2: 9added to given number results the reverse of the given number
10y + x = (10x + y )+ 9
10y + x - 10x + y = 9
9y - 9x = 9
(divide the whole equation by 9)
we get, y - x = 1 ------> 2

adding equation 1 and 2 we get ,
y + x = 9
y - x = 1
---------------
2y = 10
y= 10/2
y = 5
if y = 5 then x + y =9
x +5=9
x = 9-5
x = 4
Therfore the given number is 45...

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