Question

the sum of the digit of a 2-digit is 9 . if the digit are reversed, the new number increased by 27 . find the number​

Answer 1

Answer:

Let the unit place digit of a two-digit number be x.

Therefore, the tens place digit = 9-x

\because∵ 2-digit number = 10 x tens place digit + unit place digit

\therefore∴ Original number = 10(9-x)+x

According to the question, New number

= Original number + 27

\Rightarrow10x+\left(9-x\right)=10\left(9-x\right)+x+27⇒10x+(9−x)=10(9−x)+x+27

\Rightarrow10+9-x=90-10x+x+27⇒10+9−x=90−10x+x+27

\Rightarrow9x+9=117-9x⇒9x+9=117−9x

\Rightarrow9x+9x=117-9⇒9x+9x=117−9

\Rightarrow18x=108⇒18x=108

\Rightarrow x=\frac{108}{18}=6⇒x=

18

108

=6

Hence, the 2-digit number = 10(9-x)+x = 10(9-6)+6 = 10 x 3 + 6 = 30 + 6 = 36

Answer 2

Answer:

The number is 36

Step-by-step explanation:

Let the number be 10x+y

ATQ,

x + y = 9 (1)

also, 10y + x = 10x + y + 27

=> 9(y-x) =27

=> y - x = 27/9 = 3 (2)

(1) + (2)

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

=> 2y = 12

=> y = 12/2 = 6

So in (1)

x + 6 = 9

x = 9-6 = 3

so the number is -

10x + y

10(3) + 6

= 30 + 6 = 36

Note

The number is not taken as xy, because it means x × y. So x = 3, y = 6 would make 3×6 = 18 and not 36, So the number is taken as 10x + y as 'x' is at 10ths place and the digits of the number are 'x' and 'y'