Question

The sum of the no of a 2 digit no is 9. when the digits are reversed they differ by 27. what is the number?​


plz answer

Answer 1

Solution :

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

The sum of the number of two digit number is 9 when the digits are reversed they differ by 27.

$\bf{\blue{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

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

Let the ten's digit number be r

Let the one's digit number be m

$\underline{\sf{The\:original\:number=10r+m}}}}\\\underline{\sf{The\:reversed\:number=10m+r}}}}$

So;

$\mapsto\sf{r+m=9}\\\\\mapsto\bf{r=9-m.......................(1)}$

&

$\mapsto\sf{10r+m-(10m+r)=27}\\\\\mapsto\sf{10r+m-10m-r=27}\\\\\mapsto\sf{10r-r+m-10m=27}\\\\\mapsto\sf{9r-9m=27}\\\\\mapsto\sf{9(r-m)=27}\\\\\mapsto\sf{r-m=\cancel{\dfrac{27}{9} }}\\\\\mapsto\sf{r-m=3}\\\\\mapsto\sf{9-m-m=3\:\:\:\:[from(1)]}\\\\\mapsto\sf{9-2m=3}\\\\\mapsto\sf{-2m=3-9}\\\\\mapsto\sf{-2m=-6}\\\\\mapsto\sf{m=\cancel{\dfrac{-6}{-2} }}\\\\\mapsto\sf{\red{m=3}}$

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

$\mapsto\sf{r=9-3}\\\\\mapsto\sf{\red{r=6}}$

Thus;

The original number = 10r+m

The original number = 10(6)+3

The original number = 60 + 3

The original number = 63

Answer 2

Step-by-step explanation:

Given -

• sum of the number of two digit is 9
• when the digits are reversed they differ by 27

To Find -

What is the number ?

Let the ten's digit number be x

and

the one's digit number be y

Then,

The original number is 10x + y

and

The reversed number is 10y + x

Now,

• x + y = 9 ....... (a)

and

(10x + y) - (10y + x) = 27

= 10x + y - 10y - x = 27

= 9x - 9y = 27

= 9(x - y) = 27

• = x - y = 3 ........ (b)

Now,

Adding (a) and (b)

x + y = 9

x - y = 3

_________

2x = 12

x = 12/2

• = x = 6

substituting the value of x on (a), we get :

x + y = 9

6 + y = 9

y = 9 - 6

• = y = 3

Then,

The original number is 10x + y

= 10(6) + 3

= 60 + 3

= 63

Hence,

The required number is 63.