Question

Nine time a two digit number is the same as twice the number obtained byinterchanging the digits of the number if one digit of the number exceeds the other number by 7 find the number

Answer 1

Let the ten's digit be x and

the unit's digit be y

therefore, the number becomes = 10x+y and

when the number is reversed = 10y+x

Now, when nine times a two-digit number is the same as twice the number obtained by interchanging the digits of the number.

⇒9(10x+y) = 2 (10y+x)

⇒90x+9y = 20y+2x

⇒90x-2x = 20y - 9y

⇒88x=11y

or y = 8x.........(i)



Now, if one digit of the no. exceeds the other by 7, then

y = x+7

⇒8x = x+7 [From (i)]

⇒8x-x=7

⇒7x=7

⇒x=7/7

⇒x=1

From (i) , we get,

y = 8(1)

⇒y = 8



Thus, the number becomes = 10x+y = 10(1)+8=10+8 = 18

Answer 2

here the answer.. hope it is helpful