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

Step 1: Finding the relationship between unit digit and ten's digit of the number according to the question:

Let the unit digit of the two-digit number be$"x"$

the tens digit of a two-digit number be$"y"$

Then the number$\text{= 10y+x}$

After interchanging the digits, the number$\text{= 10x+y}$

According to the question,

$$\begin{gathered} \text{10y+x+10x+y = 88} \\ \text{11y+11x = 88} \\ \text{x+y = 88÷11} \\ \text{x+y = 8 --------------(i)} \end{gathered}$$

According to the question,

$\text{x = 3y -------------(ii)}$

Step 2: Finding the value of$\text{x and y}$:

Let's substitute the value of$x=3y$in equation (i) -

$$\begin{gathered} \text{3y+y = 8} \\ \text{4y = 8} \\ \text{y = 2} \end{gathered}$$

Substituting the value of$y=2$in equation (ii) -

$$\begin{gathered} \text{x = 3×2} \\ \text{x = 6} \end{gathered}$$

Step 3: Finding the original number:

Therefore, the original number,

$$\begin{gathered} =10y+x \\ =10\times 2+6 \\ =20+6=26 \end{gathered}$$

Hence, the original number is$26.$