Question

A two-digit number is 3 more than 4 times the sum of its digits.If 18 is added to the number, its digits are reversed.Find the number.

Answer 1

Here is your 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

Hope it helps you

Answer 2

$\huge \bf{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$


let the digit at once digit be x

and tens place be y

then

10y+x

when digit reversed

it becomes

10x+y

case 1
_______


10y+x=4(x+y)+3

10y+x=4x+4y+3

10y+x-4x-4y=3

6y-3x=3....... ( both side ÷ by 3)

2y-x=1________(1)


case 2
______

(10y+x)+18=10x+y

10y+x-10x-y=18

9x-9y=18......... ( ÷ both side by 9)

x-y=2________(2)



x=2+y

put the value of x in equation (1)


2y-X=1

2y-(2+y)=1

2y-2-y=1

y=3

put value of y in equation 2


x-y=2

x-3=2

x=5

then the number is


10y+x= 10*3+5

=35

$\huge \boxed{ \boxed{ \boxed{hope \: it \: helps}}}$

$\huge{ \mathbb{THANKS}}$