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?( class 8th question)​

Answer 1

Answer:

The original number is 62.

Step-by-step explanation:

Consider the -

The digit in units place as x .

The digit in tens place as 3x.

So,

The number is 10(3x) + x = 31x

As, given Digits are interchanged.

So,

Digit in units place will be 3x

Digit in tens place will be x

Then,

The new number :

$\sf \implies 10(x) + 3x = 13x$

As given,

Sum of these two numbers = 88.

So,

⇒ 31x + 13x = 88

⇒ 44x = 88

⇒ x = 2

⇒ 3x = 6

⇒ 62.

∴ The orignal number is 62.

Answer 2

Answer:

$\large\bold\red{62}$

Step-by-step explanation:

Let the digit at ones place be 'x'

Therefore,

The digit at tens place will be ' 3x'

Therefore,

the number = 10 × 3x + x = 30x + x = 31x

Now,

The digits are interchanged,

therefore,

the digit at ones place = 3x

the digit at tens place = x

Therefore,

the new number = 10x + 3x = 13x

According to Question,

31x + 13x = 88

=> 44x = 88

=> x = 88/44

=> x = 2

Therefore,

the original number = 31 × 2 = 62

Hence,

Required number = 62