Question

a two digitis 3 more than 4 times the sum of its digits.If 18 is added to the number ,tge digits are reversed.Find the number

Answer 1

let two digit be 10x + y and it's reversed digit be 10y + x
the two digit (10x + y) 3 more than 4 times the sum of it's digits
so, 10x + y = 3 + 4(x + y)
10x + y = 3 + 4x + 4y
10x - 4x = 3 + 4y - y
6x = 3 + 3y
divide both sides by 3
2x = 1 + y..........(1)
18 is added to the two digit (10x + y) number get reversed
so, 10x + y + 18 = 10y + x
10x - x + 18 = 10y - y
9x + 18 = 9y
divide both sides by 9
x + 2 = y
y = x + 2.................(2)
substitute (2) in (1)
2x = 1 + x + 2
x = 3
then, y = 3 + 2 = 5
so the two digit number is 10x + y = 10*3 + 5 = 30+5 = 35

Answer 2

solutions :-

Let ones place digit be x .

Tens place digit be y .

Original number = 10y + x

Number formed by reversing the digits = 10x + y

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

A/q

=>10y + x = 4(x + y) + 3 
=>10y + x − 4x − 4y = 3
 => 6y − 3x = 3 
=>2y − x = 1..................(1) 

Again when 18 is added to the number then digits get interchanged. 

(10y + x) + 18 = 10x + y 

=>9x − 9y = 18
=> x − y = 2...................(2) 

On adding equation (1) and (2):-

 2y − x +(x-y)= 1+2
2y-x+x-y = 3
Y = 3

Putting the value of y im equation (2)

=> x − y = 2
=>x- 3 =2
=>x= 2+3
=>x =5

Hence

Number = 10y +x= 10×3+5=35

Thanks