Question

3. The digit in the ten's place in a two-digit number is three times the digit in the unit's place. If the digits are reversed, the new number is 54 less than the original number. Find the original number.​

Answer 1

Answer:

I have done it with 36 so I directly sender you the formula

hope it helps

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Step-by-step explanation:

Let the one's digit be y and tens digit be x,

Number = 10x + y

Then,x=3y⋯(i)

Reversed number = 10y + x

A.t.Q :- (10x+y)−(10y+x)=36 Put x = 3y in eq. (i)

⇒9x−9y=36

⇒x−y=4⋯(ii)

⇒3y−y=4

∴2y=4 x=3y ∴x=6

y=2

∴ Number = 62

Answer 2

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

Given that,

The digit in the ten's place in a two-digit number is three times the digit in the unit's place.

Let assume that

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{digit \: at \: ones \: place \: be \: x} \\ \\ &\sf{digits \: at \: tens \: place \: be \: 3x} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 10 \times 3x + x \times 1 = 31x} \\ \\ &\sf{reverse \: number = 10 \times x + 3x \times 1 = 13x} \end{cases}\end{gathered}\end{gathered}$

Further given that,

If the digits are reversed, the new number is 54 less than the original number.

$\rm\implies \:31x - 13x = 54$

$\rm\implies \:18x = 54$

$\rm\implies \:x = 3$

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{digit \: at \: ones \: place \: be \: 3} \\ \\ &\sf{digits \: at \: tens \: place \: be \: 9} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 31x = 93} \\ \\ &\sf{reverse \: number = 13x = 39} \end{cases}\end{gathered}\end{gathered}$

Hence,

• Original number is 93.