Find the sum of 31.489 and the number got by inter changing 1 by 10 th place and 1 by 100 th place of this number​

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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

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