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 original number
Answers
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:
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: