Question

a number consists of two digits whose sum is 9 . if 27 is subtracted from the number ,the digits interchange their places find the number​

Answer 1

Answer:

Let the once place be x and tens place be y

Now x+y = 9 (given)

or x=9-y

ATQ yx-27=xy

or y*9-x-27=9-y*y

or 9y-9x-27=9y-y²

or y²-9x-27=0

At last solving equation you will get answer

Answer 2

S O L U T I O N :

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

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

$\mapsto\tt{r+m=9}$

$\mapsto\tt{r=9 - m............(1)}$

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

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

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

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

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

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

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

$\mapsto\tt{m-(9-m)=-3\:\:\:[from(1)]}$

$\mapsto\tt{m-9+m =-3}$

$\mapsto\tt{2m -9 =-3}$

$\mapsto\tt{2m=-3+9}$

$\mapsto\tt{2m=6}$

$\mapsto\tt{m=\cancel{6/2}}$

$\mapsto\bf{m=3}$

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

$\mapsto\tt{r=9-3}$

$\mapsto\bf{r=6}$

Thus;

→ The number = 10r + m

→ The number = 10(6) + 3

→ The number = 60 + 3

→ The number = 63 .