Question

the sum of the digits of a two digit number is9 .if thedigits are reversed the number is increased by 45.Find the original number

Answer 1

let the 1st and 2nd digits be x and y
x+y=9...................................................1
original no.= 10x+y.............................$
reversed no.=10y+x
(10y+x)-(10x+y)=45
9(y-x)=45
(y-x)=45/9=5.........................................................................2
now adding 1 and 2,
we get, 2y=14,
y=7.
Now, putting y=7 in 2,
7-x=5
x=2.
now according to $                                             (i.e., 10X2+7)
we have, original no.=27

Answer 2

Let the digit in Tens place = x
The digit in units place = y
The original number is 10x + y

Sum of the digits = 9
⇒x + y = 9 ....... (1)

Digits are reversed. 

Tens digit = y
Units digit = x
The new number formed = 10y + x

New number is 45 more than the old

⇒10y + x = 10x + y + 45
⇒ 9y - 9x = 45
⇒ y - x = 5      ......(2)

Adding equations (1) and (2), we get

2y = 14 ⇒ y = 7

Substituting y = 7 in equation (1) and solving, we get

x = 2

The original number is = 10x + y = 10(2) + 7 = 27

Number after reversing digits = 72

The difference = 72 - 27 = 45