Question

On reversing the digits of a two digit number obtained is 9 less than three times the original number. If the difference of these two numbers is 45 find the original number

Answer 1

let the ones place digit be y & tens place digit be x. 
the original number = 10x+y 
reversed number = 10y+x 

according to question, 
a) 3(10x+y)-9 = 10y+x 
=> 30x+3y-9-10y-x= 0 
=> 29x-7y-9 = 0-----(1) 

b) as the reversed number is bigger, so 
10y+x-(10x+y)=45 
=>10y-y-10x+x-45 = 0 
=> 9y-9x-45=0 
=> y-x-5=0 
=> y= x+5 -----(2) 

now put equation (2) in equation (1) 

29x-7(x+5)-9=0 
=> 29x-7x-35-9=0 
=> 22x=44 
=> x = 2 

now put the x=2 in equation (2) 
y= 2+5= 7 

therefore the original number is 27 & the reversed number is 72. 

Answer 2

Step-by-step explanation:

The number is 27.

a+b=9⟹a=9−b

10a+b+45=10b+a

⟹10(9−b)+b+45=10b+9−b

⟹90−10b+b+45=9b+9

⟹(135−9b)+9b=(9b+9)+9b

⟹(18b+9)−9=135−9

⟹18b18=12618

→b=7 and a=2

answer:27

Proof:

2+7=9