# Ex.2.4 Q4 Linear Equations in One Variable Solution - NCERT Maths Class 8

## Question

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

## Text Solution

**What is known?**

i) One of the two digits of a two-digit number is three times the other digit

ii) Interchanging the digit of this two-digit number and adding the resulting number to the original number results in \(88\).

**What is unknown?**

Original number

**Reasoning:**

Assume one of the digits of two-digit as variable then use other conditions and form a linear equation.

**Steps:**

Let the digits at tens place and ones place be \(x\) and \(3x\) respectively.

Therefore, original number \( = 10x + 3x = 13x\)

On interchanging the digits, the digits at ones place and tens place will be \(x\) and \(3x\)* *respectively.

Number after interchanging \( = 10 \times 3x + x = 30x + x = 31x\)

According to the given question,

Original number \(+\) New number \(=\) \(88\)

\[\begin{align}{13x + 31x = 88} \\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,44x = 88}\end{align}\]

Dividing both sides by \(44\), we obtain

\[x = 2\]

Therefore, original number \( = {\text{ }}13x = 13 \times 2 = 26\)

By considering the tens place and ones place as 3\(x\) and \(x\) respectively, the two-digit number obtained is \(62.\)

Therefore, the two-digit number may be \(26\) or \(62.\)