Question

sum of the digig of a two digit no. is 15 if the no. formed by the digits is less than the original no. by 27 . find the original no.​

Answer 1

Solution :

$\underline{\bf{Given\::}}$

The sum of the digit of a two digit number is 15. If the number formed by the digits is less than the original number by 27 .

$\underline{\bf{Explanation\::}}$

Let the ten's place digit be r & let the one's place digit be m respectively;

$\boxed{\bf{Original\:number=10r+m}}$

$\boxed{\bf{Reversed\:number=10m+r}}$

A/q

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

$\mapsto\tt{r+m=15}$

$\mapsto\tt{r=15-m..................(1)}$

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

$\mapsto\tt{(Reversed\:number) =(Original\:number) -27}$

$\mapsto\tt{10m+r= 10r+m-27}$

$\mapsto\tt{10m-m +r-10r = -27}$

$\mapsto\tt{9m-9r= -27}$

$\mapsto\tt{9(m-r)= -27}$

$\mapsto\tt{m-r= -\cancel{27/9}}$

$\mapsto\tt{m-r=-3}$

$\mapsto\tt{m-(15-m)=-3\:\:\:[from(1)]}$

$\mapsto\tt{m-15 + m=-3}$

$\mapsto\tt{2m-15=-3}$

$\mapsto\tt{2m=-3 + 15}$

$\mapsto\tt{m=\cancel{12/2}}$

$\mapsto\bf{m=6}$

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

$\mapsto\tt{r=15-6}$

$\mapsto\bf{r=9}$

Thus;

The original number = 10r + m

The original number = 10(9) + 6

The original number = 90 + 6

The original number = 96