Question

The sum of the digits of a two digit number is 12. The number formed by interchanging the digits is greater than the original by 54. Find the original numbe​r​

Answer 1

Answer:

The Orignal Number is 39

Step-by-step explanation:

Solution :

Sum of digits = 12

Number formed by interchanging the digits is greater than the original by 54

Let the digits be -

• Tens place = 10(y)
• Units place = (12 - y)

⇒ 10(y) + (12 - y)

⇒ 10y + 12 - y

⇒ 9y + 12 -----(Original number)

$\rule{300}{1.5}$

Inversed number,

• Tens place = 10(12 - y)
• Units place = y

⇒ 120 - 10y + y

⇒ 120 - 9y

$\rule{300}{1.5}$

The number formed by interchanging the digits is greater than the original by 54

⇒ (9y + 12) + 54 = 120 - 9y

⇒ 9y + 66 = 120 - 9y

⇒ 9y + 9y = 120 - 66

⇒ 18y = 54

⇒ y = 54/18

⇒ y = 3

$\rule{300}{1.5}$

Orignal Number -

⇒ 9 × 3 + 12

⇒ 27 + 12

⇒ 39

Therefore, the Orignal Number is 39.

Answer 2

Given :-

✧ Sum of digits of two digit number = 12

✧ Number after interchanging digits = 54 greater than original digit.

Solution :-

Let's assume that the original digit →

y → ones place.

x → tens place.

⇝ x + y = 12

⇝ y = 12- x

Tens place = 10x

Ones place = 12-x

⇝Original Digit = 10x + (12-x)

⇝ 10x + 12 - x

⇝ 9x +12. ........... eq 1st

Now after interchanging the digits →

Ones → x

Tens place → 10( 12-x)

⇝ 10(12-x) + x

⇝ 120 - 10x +x

⇝ 120 -9x .............eq 2nd

According to the question eq 2nd is 54 digits greater than 1st

⇝ 120 - 9x = 54 + 9x + 12

⇝ 120 - 54 - 12 = 9x + 9x

⇝ 120 - 66 = 18x

⇝ 54 = 18x

⇝ x = 3

Original number = 10x + ( 12-x)

→ 10(3) + (12-3)

→ 30 + 9

→ 39 is the required number .

On interchanging the digit

93 = 54 + 39