Question

The sum of a two digit number is 11 if the digits at the tens place is incresed by 5 and the digits unit place is decreased by 5 the digits of the number are found to be reversed find the original number

Answer 1

Answer:

here is your answer;-)

Step-by-step explanation:

Let us assume, x is a tenth place digit and y is the unit place digit of a two-digit number.

Therefore, the number = 10x + y and the reversed number = 10y + x

Given:

x + y = 11 -----------1

Also given:

10(x + 5) + y - 5 = 10y + x

10x + 50 + y - 5 = 10y + x

10x + y + 45 = 10y + x

9y - 9x = 45

y - x = 5 ---------------2

Adding equation 1 and equation 2

2y = 16

y = 8

Therefore, x = 11 - y = 11 - 8 = 3

Therefore, the two-digit number = 10x + y = 10*3 + 8 = 38

Answer 2

Solution :

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

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

A/q

$\longrightarrow\sf{r+m=11}\\\\\longrightarrow\sf{r=11-m.......................(1)}$

&

$\longrightarrow\sf{10m+r=10(r+5)+(m-5)}\\\\\longrightarrow\sf{10m+r=10r+50+m-5}\\\\\longrightarrow\sf{10m+r=10r+m+45}\\\\\longrightarrow\sf{10m-m+r-10r=45}\\\\\longrightarrow\sf{9m-9r=45}\\\\\longrightarrow\sf{9(m-r)=45}\\\\\longrightarrow\sf{m-r=\cancel{45/9}}\\\\\longrightarrow\sf{m-r=5}\\\\\longrightarrow\sf{m-(11-m)=5\:\:[from(1)]}\\\\\longrightarrow\sf{m-11+m=5}\\\\\longrightarrow\sf{2m-11=5}\\\\\longrightarrow\sf{2m=5+11}\\\\\longrightarrow\sf{2m=16}$

$\longrightarrow\sf{m=\cancel{16/2}}\\\\\longrightarrow\bf{m=8}$

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

$\longrightarrow\sf{r=11-8}\\\\\longrightarrow\bf{r=3}$

Thus;

$\boxed{\sf{The\:Original\:number\:=10r+m=[10(3)+8]=[30+8]=\boxed{\bf{38}}}}}$