Question

sum of the digits of a two-digit number is 9 if the number obtained by reversing the order of digits is 27 more than the original number find the original number ​

Answer 1

Answer:

$therefore \: the \: original \: number \: is \: \: 36.$

Step-by-step explanation:

$let \: the \: unit \: digit \: is \: u \\ \\ and \: the \: tens \: digit \: be \: t \\ \\ then \\ \\ u + t = 9 \: \: \: \: ..(i) \\ \\ t + 10u = 27 + 10t + u \\ \\ 9u - 9t = 27 \\ \\ u - t = 3 \: \: \: \: ...(ii)\\ \\ solving \: these \\ \: we \: get \\ \\ 2u = 12 \\ \\ u = 6 \\ \: \: then \: \: t = 3 \\ \\ therefore \: the \: original \: number \: is \: \: 36.$

Answer 2

Answer: Answer is 36 as 3+6=9. If we reverse it we will get 63. and 36+27 is also 63.

Step-by-step explanation:

Let the two digit number be 10x+y.

Given that the sum of the digits is 9

x+y=9     (1)

Given that the number obtained by interchanging the digits exceeds the given number by 27

10y+x=10x+y+27

9x−9y=−27

On taking 9 as common

x−y=−3           (2)

Adding equation (1) and (2)

x+y=9

x−y=−3

2x=6

x=3

3+y=9

y=6

Therefore, the number is 10x+y is 36.

Hence, this is the answer.

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