Question

If a two digit no is 3 more than 4 times the sum of digits and if 18 is added to the no, digits are reversed . Find the no.

Answer 1

Let the digit in one's place be x and the digit in ten's place be be y.

Hence, the original number = 10x + y.

Given that A two digit number is 3 more than 4 times the sum of digits.

⇒ 10x + y = 3 + 4(x + y)

⇒ 10x + y = 3 + 4x + 4y

⇒ 6x - 3y = 3

⇒ 2x - y = 1    ----- (1)

Given that if 18 is added to the original number, then the digits are reversed.

⇒ 10x + y + 18 = 10y + x

⇒ 9x - 9y = -18

⇒ x - y = -2     ------ (2)

On solving (1) & (2), we get

⇒ 2x - y = 1

⇒ x - y = -2

   ----------------

    x = 3

Substitute x = 3 in (1), we get

⇒ 2x - y = 1

⇒ 2(3) - y = 1

⇒ 6 - y = 1

⇒ y = 5.

Hence, the original number :

⇒ 10x + y

⇒ 10(3) + 5

⇒ 35.

Therefore, the original number = 35.

Hope this helps!

Answer 2

Here is your solution

Let,

The digit in ones place be x

Tens place be y Original number = 10y+x

Number formed by reversing the digits = 10+y

Given:-
10y + x = 4(x + y) + 3 
⇒ 10y + x − 4x − 4y = 3 
⇒ 6y− 3x= 3 
⇒ 2y − x= 1 ..........(1)

Also,
Given that when 18 is added to the number the digits gets interchanged.

Hence

(10y + x) + 18 = 10x + y
 ⇒ 9x − 9y = 18
 ⇒ x − y = 2 ....... (2) 

Adding (1) and (2),

we get
2y− x+x-y = 1+2
 y = 3

Put y=3
in x − y = 2,
we get 
x − 3 = 2
 ∴ x = 5

Original Number = (10y +x)
= (10×3+ 5 )
= 35