Question

The 2 digit number which becomes 5/6 th
of itself when its digits are reversed. If
the difference in the digits of the number being 1, what is the two digits number?

Answer 1

Solution :-

Let the unit place digit be x and tens place digit be y

Therefore,

Original number = 10y + x

After reversing the digits,

New Number = 10x + y

According to the question,

The 2 digit number which becomes 5/6th of itself when it's digits are reversed.

( 10y + x) = 5 /6 ( 10x + y)

5( 10y + x) = 6( 10x + y)

50y + 5x = 60x + 6y

50y - 6y = 60x - 5x

44y = 55x

44y - 55x = 0

11( 4y - 5x) = 0

4y - 5x = 0. eqn ( 1 )

Now,

The difference in the digits

x - y = 1

x = 1 + y eqn ( 2 )

Subsitute eqn( 2 ) in eqn( 1 )

4y - 5( 1 + y) = 0

4y - 5 + y = 0

5y = 5

y = 5/5

y = 1

Now, Subsitute the value of y in eqn ( 2 )

x = 1 + 1

x = 2

Hence, The two digit is 12 $.$

Answer 2

Answer:

$\huge \mathfrak \purple{Solution}$

Let the unit place digit be x and tens place digit be y

Therefore,

Original number = 10y + x

After reversing the digits,

New Number = 10x + y

According to the question,

The 2 digit number which becomes 5/6th of itself when it's digits are reversed.

( 10y + x) = 5 /6 ( 10x + y)

5( 10y + x) = 6( 10x + y)

50y + 5x = 60x + 6y

50y - 6y = 60x - 5x

44y = 55x

44y - 55x = 0

11( 4y - 5x) = 0

4y - 5x = 0. eqn ( 1 )

Now,

The difference in the digits

x - y = 1

x = 1 + y eqn ( 2 )

Subsitute eqn( 2 ) in eqn( 1 )

4y - 5( 1 + y) = 0

4y - 5 + y = 0

5y = 5

y = 5/5

y = 1

Now, Subsitute the value of y in eqn ( 2 )

x = 1 + 1

x = 2

▶Hence, The two digit is 12