Question

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Answer 1

let the ones digit place be x then the tens digit place will be 10 (9-x)

new digit 10×x +9-x

we have,

10x+9-x - 10(9-x)+x=27

9x+9-90+9x=27

18x=27+90-9

18x=108

x=108/18=6

the two digits are 36.

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Answer 2

Answer :

36

Solution :

Let the tens digit and the unit digit of the required two digits number be x and y respectively .

Then ,

Required (Original) number = 10x + y

Also , it is given that the sum of digits of the required number is 9 .

ie. x + y = 9 --------(1)

Now ,

On interchanging the digits of the original number , the tens digit and the unit digit of the new number will be y and x respectively .

Then ,

New number = 10y + x

Now ,

According to the question , on interchanging the digits of original number , the resulting number is greater than original number by 27 .

Thus ,

=> 10y + x = 10x + y + 27

=> 10y + x - 10x - y = 27

=> 9y - 9x = 27

=> 9(y - x) = 27

=> y - x = 27/9

=> y - x = 3 --------(2)

Now ,

Adding eq-(1) and (2) , we get ;

=> x + y + y - x = 9 + 3

=> 2y = 12

=> y = 12/2

=> y = 6

Now ,

Using eq-(1) , we have ;

=> x + y = 9

=> x + 6 = 9

=> x = 9 - 6

=> x = 3

Hence,

The required number is 36 .

Alternative :

By inspection (not to be used in subjective exams) -

Here it is given that , the sum digits of a two digits number is 9 .

And on interchanging the digits of original number , the resulting number is greater than original number by 27 .

(ie. new number > original number)

Now ,

Let's write all the possible two digits numbers whose sum of digits is 9 and whose interchanged number is greater than the original number .

Original no. New no. after

interchanging the digits

18 81

27 72

36 63

45 54

Clearly ,

For the number 36 , the new number obtained on interchanging the digits is 63 which is greater than 36 by 27 .

Hence ,

Required number is 36 .