Question

The sum of digit of a two digit number is 15,if the original number formed by changing the place of the digit is greater than the orginally number by 9,find The original number​

Answer 1

Let the digit at the unit place be x.

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9x

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9xOn reversing the digits, we have x at the tens place and (15−x) at the unit place.

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9xOn reversing the digits, we have x at the tens place and (15−x) at the unit place.∴ New number =10x+(15−x)=(9x+15)

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9xOn reversing the digits, we have x at the tens place and (15−x) at the unit place.∴ New number =10x+(15−x)=(9x+15)According to the given condition,

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9xOn reversing the digits, we have x at the tens place and (15−x) at the unit place.∴ New number =10x+(15−x)=(9x+15)According to the given condition,(Original number) − (New number) =27

Let the digit at the unit place be x.Then, the digit in the tens place =(15−x)∴ Original number =10×(15−x)+x=150−9xOn reversing the digits, we have x at the tens place and (15−x) at the unit place.∴ New number =10x+(15−x)=(9x+15)According to the given condition,(Original number) − (New number) =27⇒(150−9x)−(9x+15)=27⇒150−9x−9x−15=27⇒135−18x=27⇒18x=135−27⇒x=(18108)=6