Question

A two digit number is 3 more than 4 times the sum of its digits. if 18 is subtracted from the number the digits are reversed find the number

Answer 1

A digit number=10x+y,here two numbers are x,y
when two digit number is 3 more than 4 times the sum of digits =4(x+y)=10x+y+3
4x+4y=10x+y+3
10x-4x+y-4y+3=0
6x-3y+3=0.....(I)
when 18 is substacted from number the digits are reversed=10x+y-18=10y+x
10x-x+y-10y-18=0
9x-9y=18
9(x-y)=18
x-y=2....(ii)
x=2+y
put x value in equ (I)
6x-3y+3=0
6(2+y)-3y+3=0
12+6y-3y+3=0
3y=-15
y=-5
put y value in equ (ii)
x-y=2
x=2+y
x=2-5
x=-3
ans: 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