Question

The ones digits of a 2-digit number is twice the tens digits. When the number formed by reversing the digits is added to the original number, the sum is 999. find the original number
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Answer 1

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Let the digits at tens place and ones place: x and 9−x respectively.

∴ original number =10x+(9−x)

=9x+9

Now Interchange the digits: Digit at ones place and tens place: x and 9−x respectively.

∴ New number: 10(9−x)+x

=90−10x+x

=90−9x

AS per the question

New number = Original number +27

90−9x=9x+9+27

90−9x=9x+36

18x=54

x= 18/54

x=3

Digit at tens place ⇒3 and one's place : 6

∴ Two digit number: 36

Answer 2

Let the digits at tens place and ones place: x and 9−x respectively.

∴ original number =10x+(9−x)

=9x+9

Now Interchange the digits: Digit at ones place and tens place: x and 9−x respectively.

∴ New number: 10(9−x)+x

=90−10x+x

=90−9x

AS per the question

New number = Original number +27

90−9x=9x+9+27

90−9x=9x+36

18x=54

x= 18/54

x=3

Digit at tens place ⇒3 and one's place : 6

∴ Two digit number: 36