It is given that sum of the digits of the two digit number is 9.

Let, ones place digit bexxand tens place digit be9−x9−x.

So the original number is,

{x+10×(9−x)}=90−9x{x+10×(9−x)}=90−9x

After interchanging the places of digits new number will be,

(9−x+10x)=9+9x(9−x+10x)=9+9x

New number is 27 greater than the original number. So,

(9+9x)=(90−9x)+279x+9x=27+90−918x=108x=6(9+9x)=(90−9x)+279x+9x=27+90−918x=108x=6

Original number is,

{x+10×(9−x)}=6+10(9−6)=6+10(3)=36{x+10×(9−x)}=6+10(9−6)=6+10(3)=36

Therefore, the original two digit number is 36.

let the ones digit be x and tens be ynow number formed = 10y+xaccording to the conditions given in the question,. x + y = 9 ... ( i) 10x +y = 10y +x + 279x - 9y = 27x - y = 3 ... (ii) adding eq i and ii, 2x = 12x = 6and y = 3number = 10y +x =10x3 + 6number = 36 ans

let the ones digit be x and tens be ynow number formed = 10y+xaccording to the conditions given in the question,. x + y = 9 ... ( i) 10x +y = 10y +x + 279x - 9y = 27x - y = 3 ... (ii) adding eq i and ii, 2x = 12x = 6and y = 3number = 10y +x =10x3 + 6number = 36 ans

36 is the answer........

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36 is the correct answer of the given question

36 is the correct answer