Question

The digit at ones place of a 2-digit number is four times the digit at tens place. The number obtained by reversing the digits exceeds the given number by 54. Find the given number.​

Answer 1

Let the tens digit be x and the ones digit be y

y = 4x

Original number = 10x + y

= 10x + 4x

= 14x

Reversed number = 10y + x

= 10 (4x) + x

= 41x

Now, 41x - 14x = 54

27x = 54

x = 2

y = 4x = 4 (2) = 8

Original number = 10x + y

= 10 (2) + 8

= 28

Answer 2

Answer:

Let

The digit at ten's place be x

Then, digit at ones place = 4x

Original number = 10x + 4x = 14x

On reversing the digits,

New number = 10(4x) + x

= 40x + x

=41x

ATQ

= 41x - 14x = 54

= 27x = 54

= x = 54/27

= x = 2

Therefore , the given number = 14x

=14×2

=28