Question

The sum of the digits of a two-digit number is 9 . Also , nine times this number is twice the number obtained by reversing the order of the digit . Find the number by cross-multiplication method .​

Answer 1

Answer:

Let the ten's digit no. be x and one's digit no. be y.

Let the ten's digit no. be x and one's digit no. be y.So the no. will be = 10x+y.

Let the ten's digit no. be x and one's digit no. be y.So the no. will be = 10x+y.Given :  x+y=9-----(I)

Let the ten's digit no. be x and one's digit no. be y.So the no. will be = 10x+y.Given :  x+y=9-----(I)             9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)

Let the ten's digit no. be x and one's digit no. be y.So the no. will be = 10x+y.Given :  x+y=9-----(I)             9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)On solving I and II simultaneously you will get x=1 and y=8.

Let the ten's digit no. be x and one's digit no. be y.So the no. will be = 10x+y.Given :  x+y=9-----(I)             9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)On solving I and II simultaneously you will get x=1 and y=8.Therefore your desired no. is 18