Question

the sum of all the digits of a 2 digit number is 9. the digits are reserved and the new numbers is subtracted from the original number, to get 45. find the original number. ​

Answer 1

Step-by-step explanation:

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

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