# 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

2
by axshat

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by axshat

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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

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