Question

One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? ​

Answer 1

Let the digit at tens place be x

The digit at ones place will be 3x.

Original two-digit number = 10x + 3x

After interchanging the digits, the new number = 30x + x

According to the given details, the equation becomes  

(30x + x) + (10x + 3x) = 88

Rearranging the equation and combining like terms we get

⇒ 31x + 13x = 88

⇒ 44x = 88  

⇒ x = 2

Original number = 10x + 3x = 13x = 13×2 = 26

Answer  

• Hence the number is 26

Answer 2

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

Given that,

One of the two digits of a two digit number is three times the other digit.

Two cases arises.

Case :- 1

Let assume that digit at ones place be x

So, digit at tens place be 3x

Thus,

Original two digit number = 1 × x + 10 × 3x = x + 30x = 31x

Reverse number = 10 × x + 1 × 3x = 10x + 3x = 13x

Now, According to statement

If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88.

$\rm \: 31x + 13x = 88$

$\rm \: 44x = 88$

$\rm\implies \:x = 2$

Hence,

$\rm\implies \:Original \: two \: digit \: number = 31 \times 2 = 62$

Case :- 2

Let assume that digit at tens place be x

So, digit at ones place be 3x

Thus,

Original two digit number = 10 × x + 1 × 3x = 10x + 3x = 13x

Reverse number = 1 × x + 10 × 3x = x + 30x = 31x

Now, According to statement

If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88.

$\rm \: 31x + 13x = 88$

$\rm \: 44x = 88$

$\rm\implies \:x = 2$

Hence,

$\rm\implies \:Original \: two \: digit \: number = 13 \times 2 = 26$

Thus,

$\begin{gathered}\begin{gathered}\rm\implies \:\bf\: Original \: two \: digit \: number \: is \: \begin{cases} &\sf{62} \\ \\ &\sf{or} \\ \\ &\sf{26} \end{cases}\end{gathered}\end{gathered}$