Question

the sum of the digit of a two digit number is 13. the number formed by interchanging the digits is 45 more than the original number find the original number​

Answer 1

Answer:

original number is 49

Step-by-step explanation:

let the two digits of a two digit no. be 'x' and 'y'

x+y=13

x=13-y .....(i)

let the original no.be 10x+y

and the interchangeing no. be 10y+x

ATQ,

(10x+y)+45= 10y+x

10(13-y)+y+45=10y+(13-y) .....(from (i))

130-10y +y+45=10y+13-y

175-9y=9y+13

175-13=9y+9y

162=18y

y=9

substituting value of y in equation (i)..

x=13-y

x=13-9

x=4

so the orignal no. = 10x+y

= 10(4)+(9)

= 40+9

= 49

hence, the original no. is 49.

Answer 2

S O L U T I O N :

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

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

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

A/q

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

$\mapsto\tt{x+y = 13}$

$\mapsto\tt{x = 13-y................(1)}$

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

$\mapsto\tt{(Reversed\:number) = (original\:number) + 45}$

$\mapsto\tt{(10y+x) = (10x+y) + 45}$

$\mapsto\tt{10y+x= 10x+y + 45}$

$\mapsto\tt{10y-y+x-10x= 45}$

$\mapsto\tt{9y-9x= 45}$

$\mapsto\tt{9(y-x)= 45}$

$\mapsto\tt{y-x= \cancel{45/9}}$

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

$\mapsto\tt{y-(13-y)=5\:\:[from(1)]}$

$\mapsto\tt{y-13+y=5}$

$\mapsto\tt{2y-13=5}$

$\mapsto\tt{2y=5+13}$

$\mapsto\tt{2y=18}$

$\mapsto\tt{y=\cancel{18/2}}$

$\mapsto\bf{y = 9}$

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

$\mapsto\tt{x = 13-9}$

$\mapsto\bf{x = 4}$

Thus,

The original number = 10x + y

The original number = 10(4) + 9

The original number = 40 + 9

The original number = 49