Question

The digits at the tens place of a two-digit number is four times the digit at its units place. If the digits are reversed, the new number formed will be 54 less than the original number. Find the original number.

Answer 1

Let the original two digit number be :

↣ (10y + x)

Wherein :

↣ y = Tens place

↣ x = Units place

Im the question, it is given that, the digit at the tens place is four times the digit at its units place :

↣ y = 4x . . . . . . . ( eq·1 )

Also, if the digits are reversed, the new number formed will be 54 less than the original number.

Let the reversed number be :

↣ (10x + y)

So, we'll have :

↣ (10y + x) – 54 = (10x + y)

↣ 10y + x – 54 – 10x – y = 0

↣ 9y – 9x = 54

↣ 9 (4x) – 9x = 54

↣ 36x – 9x = 54

↣ 27x = 54

↣ x = 54 / 27

↣ x = 2

Finding y from ( eq·1 ) :

↣ y = 4x

↣ y = 4 (2)

↣ y = 8

Finally, substituting these values in the original number form to get the number :

↣ (10y + x)

↣ 10 (8) + 2

↣ 80 + 2

↦ 82

━━━━━━━━━━━━━━━━━

↣Therefore, 82 is the original number.

Answer 2

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf\:\rm :\longmapsto\:Let-\begin{cases} &\sf{ones \: digit \: be \: x} \\ &\sf{tens \: digit \: be \: y} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So-\begin{cases} &\sf{number \: formed \: = \: 10y + x} \\ &\sf{reverse \: number = 10x + y} \end{cases}\end{gathered}\end{gathered}$

According to statement,

The digits at the tens place of a two-digit number is four times the digit at its units place.

$\rm :\implies\:y = 4x - - - (1)$

According to statement,

If the digits are reversed, the new number formed will be 54 less than the original number.

$\rm :\longmapsto\:10x + y = 10y + x - 54$

$\rm :\longmapsto\:10x + y - 10y - x = - 54$

$\rm :\longmapsto\:9x - 9y = - 54$

$\rm :\longmapsto\:x - y = - 6$

$\rm :\longmapsto\:x - 4x = - 6$

$\rm :\longmapsto\: - 3x = - 6$

$\bf\implies \:x = 2$

On substituting x = 2 in equation 1, we get

$\bf\implies \:y = 8$

So, it means

$\begin{gathered}\begin{gathered}\bf\:\rm :\longmapsto\:So-\begin{cases} &\sf{ones \: digit \: = \:2} \\ &\sf{tens \: digit \: = \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Hence-\begin{cases} &\sf{number \: formed \: = \: 10y + x = 82} \\ &\sf{reverse \: number = 10x + y = 28} \end{cases}\end{gathered}\end{gathered}$

So, original two digit number is 82.

Alter method :-

Let digit at ones place be x

So,

Given that

The digits at the tens place of a two-digit number is four times the digit at its units place.

It means, digit at tens place be 4x.

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So-\begin{cases} &\sf{number \: formed \: = \: 10(4x) + x = 41x} \\ &\sf{reverse \: number = 10x + 4x = 14x} \end{cases}\end{gathered}\end{gathered}$

According to statement,

If the digits are reversed, the new number formed will be 54 less than the original number.

$\rm :\longmapsto\:14x = 41x - 54$

$\rm :\longmapsto\:14x - 41x = - 54$

$\rm :\longmapsto\:- 27x = - 54$

$\bf\implies \:x = 2$

Hence,

$\begin{gathered}\begin{gathered}\bf\:\rm :\longmapsto\:So-\begin{cases} &\sf{ones \: digit \: = \:2} \\ &\sf{tens \: digit \: = \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So-\begin{cases} &\sf{number \: formed \: = \: 41x = 82} \\ &\sf{reverse \: number = 14x = 28} \end{cases}\end{gathered}\end{gathered}$

Hence,

↝ Original two-digit number is 82.