The digit at the ten's place of a two digit number is four times the digit at one's place. If the sum of this number and the number formed by reversing the digits is 55, find the numbers.

Given :

The digit at the ten's place of a two digit number is four times the digit at one's place.

The sum of the number and the number formed by reversing the digits = 55

To do :

We have to find the original number.

Solution :

 Let the two digit number be $10x+y$.

$x = 4y$

The number formed on reversing the digits is $10y+x$.

Therefore,

$10y+x + 10x+y = 55$

$(10y+y)+(x+10x) = 55$

$11(x+y) = 55$

$x+y = 5$

$4y+y = 5$

$5y = 5$

$y = 1$

$x = 4(1) = 4$

The original number is $10(4)+1 = 40+1 = 41$.

The original number is 41. 

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Updated on: 10-Oct-2022

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