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

Answer:

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

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

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

=>6x-3y=3

=>3(2x-y)=3

=>2x-y=1......(1)

10x+y+18=10y+x

=>9x-9y= -18

=>x-y= -2......(2)

Subtractibg equation (2) from (1),

2x-y-x+y=1+2

=>x=3

From (2),

x-y= -2

=>3-y= -2

=>y=5

The no. is 35.

Answer 2

S O L U T I O N :

Let the ten's digit number be x & let the one's digit be y respectively.

$\boxed{\bf{Original\:number = 10x + y}}$

$\boxed{\bf{Reversed\:number = 10y + x}}$

A/q

$\underbrace{\sf{1^{st}\:Case\::}}$

$\mapsto\tt{10x + y = 4(x+y) + 3}$

$\mapsto\tt{10x + y = 4x+4y + 3}$

$\mapsto\tt{10x -4x+ y -4y = 3}$

$\mapsto\tt{6x-3y = 3}$

$\mapsto\tt{3(2x - y) = 3}$

$\mapsto\tt{2x - y = \cancel{3/3}}$

$\mapsto\tt{2x - y = 1}$

$\mapsto\tt{2x =1 + y}$

$\mapsto\tt{x =1 + y/2...............(1)}$

$\underbrace{\sf{2^{nd}\:Case\::}}$

$\mapsto\tt{10x+y + 18 = 10y + x}$

$\mapsto\tt{10x-x+y-10y =-18}$

$\mapsto\tt{9x-9y =-18}$

$\mapsto\tt{9(x-y)=-18}$

$\mapsto\tt{x-y =- \cancel{18/9}}$

$\mapsto\tt{x - y = -2}$

$\mapsto\tt{\bigg(\dfrac{1+y}{2} \bigg) - y = -2\:\:[from(1)]}$

$\mapsto\tt{1+ y -2y = -4}$

$\mapsto\tt{1-y = -4}$

$\mapsto\tt{-y = -4-1}$

$\mapsto\tt{\cancel{-}y =\cancel{ -}5}$

$\mapsto\bf{y = 5}$

∴ Putting the value of y in equation (1),we get;

$\mapsto\tt{x =\dfrac{1+5}{2} }$

$\mapsto\tt{x =\cancel{\dfrac{6}{2} }}$

$\mapsto\bf{x =3}$

Thus,

⇒ The number = 10x + y

⇒ The number = 10(3) + 5

⇒ The number = 30 + 5

⇒ The number = 35