Question

Sum of the digits of a two digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 9. What is the two digit 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

$\huge\bold\blue{follow}$

Answer 2

Step-by-step explanation:

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