A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed.

Let the tens and the units digits of the required number be x and y, respectively. 

Required number = (10x + y) 

10x + y = 7(x + y) 

10x + 7y = 7x + 7y or 3x – 6y = 0 ……….(i) 

Number obtained on reversing its digits = (10y + x) 

(10x + y) - 27 = (10y + x) 

⇒10x – x + y – 10y = 27 

⇒9x – 9y = 27 

⇒9(x – y) = 27 

⇒x – y = 3 ……..(ii) 

On multiplying (ii) by 6, we get: 

6x – 6y = 18 ………(iii) 

On subtracting (i) from (ii), we get: 

3x = 18 

⇒ x = 6 

On substituting x = 6 in (i) we get 

3 × 6 – 6y = 0 

⇒ 18 – 6y = 0 

⇒ 6y = 18 

⇒ y = 3 

Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63 

Hence, the required number is 63.

Let unit’s digit = y

and the ten’s digit = x

So, the original number = 10x + y

The sum of the number = 10x + y

The sum of the digit = x + y

According to the question,

10x + y = 7(x + y)

⇒ 10x + y = 7x + 7y

⇒ 10x + y – 7x – 7y = 0

⇒ 3x – 6y = 0

⇒ x – 2y = 0

⇒ x = 2y …(i)

The reverse number = x + 10y

and 10x + y – 27 = x + 10y

⇒ 10x + y – 27 = x + 10y

⇒ 10x – x + y – 10y = 27

⇒ 9x – 9y = 27

⇒ x – y = 3 …(ii)

On substituting the value of x = 2y in Eq. (ii), we get

x – y = 3

⇒ 2y – y = 3

⇒ y = 3

On putting the value of y = 3 in Eq. (i), we get

x = 2(3) = 6

So, the Original number = 10x + y

= 10×6 + 3

= 60 + 3

= 63

Hence, the two digit number is 63.