Question

The sum of the digit of 2digit number is 9 interchanging the place of the digit the number reduces by 63.find the original number. ​

Answer 1

Let the number be 10x + y

X+y= 9.................1

10x+y - (10y +x) = 63

9x-9y = 63

x-y = 7...................2

Adding equation 1 and 2

we get

2x = 16

X = 8

y = 1

Number is 10x + y

Number is 10x + y 81

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Answer 2

Let, one of the digit be, 'x'.

So, the other digit = (9-x)

∴ The number is= (x×10)+9-x

                           = 10x-x+9

                           = 9x+9

By changing the place of the digits, the new number we get = (9-x)×10+x

                                                                                                     = 90-10x+x

                                                                                                     = 90-9x

By condition,

(9x+9)-(90-9x)=63 [(-)×(-) = (+)]

⇒9x+9x+9-90=63

⇒18x-81 = 63

⇒18x = 63+81

⇒18x= 144

⇒x = 144/18

⇒x=8

∴ One of the digit is 8.

So, the other digit is (9-8)

                                 = 1

∴ The number is = (8×10)+1

                           = 80+1

                          = 81

(Ans): 81