sum of digits of two digit number is 9. If we interchange the digit than sum of original number and new number is 99. Find the ori

2 thoughts on “sum of digits of two digit number is 9. If we interchange the digit than sum of original number and new number is 99. Find the ori”

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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2. Answer:

   The sum of the two digits = 9

   On interchanging the digits, the resulting new number is greater than the original number by 27.

   Let us assume the digit of units place = x

   Then the digit of tens place will be = (9 – x)

   Thus the two-digit number is 10(9 – x) + x

   Let us reverse the digit

   the number becomes 10x + (9 – x)

   As per the given condition

   10x + (9 – x) = 10(9 – x) + x + 27

   ⇒ 9x + 9 = 90 – 10x + x + 27

   ⇒ 9x + 9 = 117 – 9x

   On rearranging the terms we get,

   ⇒ 18x = 108

   ⇒ x = 6

   So the digit in units place = 6

   Digit in tens place is

   ⇒ 9 – x

   ⇒ 9 – 6

   ⇒ 3

   Hence the number is 36

   Step-by-step explanation:

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