Question

solution
12. The digit in the tens place of a two-digit number is three times that in the units place. If the
digits are reversed, the new number will be 36 less than the original number. Find the
original number. Check your solution​

Answer 1

Answer:

let units digit be x and tens digit be y

therefore no. = 10y + x

According to the ist condition of question

y = 3x (1)

On reversing the digits

units digit = y

tens digit = x

Therefore no. = y + 10x

According to the question

x + 10y = y + 10x - 36

=》x + 10y - y -10x + 36 = 0

=》 - 9x + 9y + 36 = 0 (2)

using equation 1 in 2

- 9x + 9 (3x) + 36 = 0

=》- 9x + 27x + 36 = 0

=》18x = -36

=》x = -36/18

=》 x = - 2 (3)

Using equation 3 in 1

y = 3x

=》 y = 3 (-2)

=》y = -6 (4)

Therefore x = -2

y = -6

i hope u will add this as brainlist ans.

Answer 2

S O L U T I O N :

Let the tens place digit be r

Let the ones place digit be m

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

A/q

$\longrightarrow\sf{r=3m...................(1)}$

&

$\longrightarrow\sf{10r+m-36=10m+r}\\\\\longrightarrow\sf{10r-r+m-10m=36}\\\\\longrightarrow\sf{9r-9m=36}\\\\\longrightarrow\sf{9(r-m)=36}\\\\\longrightarrow\sf{r-m=\cancel{36/9}}\\\\\longrightarrow\sf{r-m=4}\\\\\longrightarrow\sf{3m-m=4\:\:\:[from(1)]}\\\\\longrightarrow\sf{2m=4}\\\\\longrightarrow\sf{m=\cancel{4/2}}\\\\\longrightarrow\bf{m=2}$

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

$\longrightarrow\sf{r=3(2)}\\\\\longrightarrow\bf{r=6}$

Thus;

$\underbrace{\sf{The\:original\:number=(10r+m)=[10(6)+2]=60+2=\boxed{62}}}}}}$

C H E C K I N G :

10r+m - 36 = 10m + r

10(6) + 2 - 36 = 10(2) + 6

60 + 2 - 36 = 20 + 6

62 - 36 = 26

26 = 26