The ones digit of a two-digit number is half of the tens digit when the number is reversed it is 27 less than the original number. Find the original number​.

Given :

The ones digit of a two-digit number is half of the tens digit.

When the number is reversed it is 27 less than the original number.

To do :

We have to find the original number.

Solution :

 Let the two digit number be $10x+y$.

$y = \frac{1}{2}x$

$\Rightarrow x=2y$......(i)

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

Therefore,

$10y+x = (10x+y)-27$

$10y-y+x-10x = -27$

$9(y-x) = -27$

$y-x = -3$

$y-(2y) = -3$                       [From (i)]

$-y = -3$

$y = 3$

$\Rightarrow x = 2(3) = 6$

The original number is $10(6)+3 = 60+3= 63$.

The original number is 63. 

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Updated on: 10-Oct-2022

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