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 reserved find the number​

Answer 1

Let the required number be ab.

Sum of digits = a + b ........ ( 1 )

10 a + b = 4 ( a + b )

10 a + b + 18 = 10 b + a

=> 9 a + 18 = 9 b

=> a + 2 = b

10 a + a + 2 = 4 ( a + a + 2 )

=> 11 a + 2 = 8 a + 8

=> 3 a = 6

=> a = 3

=> a + 2 = 5

=>ab = 35......... [ A ]

So the number becomes 35.

ANSWER :

The original number is 35.

Answer 2

SOLUTION :

Let the ten's digit number be r

Let the one's digit number be m

$\boxed{\bf{The\:original\:number\:=10r+m}}}}}\\\boxed{\bf{The\:reversed\:number\:=10m+r}}}}}$

A/q

$\longrightarrow\sf{10r+m=4(r+m)+3}\\\\\longrightarrow\sf{10r+m=4r+4m+3}\\\\\longrightarrow\sf{10r-4r+m-4m=3}\\\\\longrightarrow\sf{6r-3m=3}\\\\\longrightarrow\sf{6r=3+3m}\\\\\longrightarrow\sf{r=3+3m/6......................(1)}$

&

$\longrightarrow\sf{10r+m+18=10m+r}\\\\\longrightarrow\sf{10r-r+m-10m=-18}\\\\\longrightarrow\sf{9r-9m=-18}\\\\\longrightarrow\sf{9(r-m)=-18}\\\\\longrightarrow\sf{r-m=\cancel{-18/9}}\\\\\longrightarrow\sf{r-m=-2}\\\\\longrightarrow\sf{\dfrac{3+3m}{6} -m=-2\:\:\:\:[from(1)]}\\\\\longrightarrow\sf{3+3m-6m=-12}\\\\\longrightarrow\sf{3-3m=-12}\\\\\longrightarrow\sf{-3m=-12-3}\\\\\longrightarrow\sf{-3m=-15}\\\\\longrightarrow\sf{m=\cancel{-15/-3}}\\\\\longrightarrow\bf{m=5}$

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

$\longrightarrow\sf{r=\dfrac{3+3(5)}{6} }\\\\\\\longrightarrow\sf{r=\dfrac{3+15}{6} }\\\\\\\longrightarrow\sf{r=\cancel{\dfrac{18}{6}} }\\\\\\\longrightarrow\bf{r=3}$

Thus;

$\underbrace{\sf{The\:number\:=10r+m=[10(3)+5]=[30+5]=\boxed{\bf{{35}}}}}}$