Question

The sum of a two-digit number and the number obtained

required by reversing its digits is 121. Find the number if it‟s

unit place digit is 5.​

Answer 1

Given units digit is x and tens digit is y

Hence the two digit number = 10y + x

Number obtained by reversing the digits = 10x + y

Given that sum of a two digit number and the number obtained by reversing the order of its digits is 121.

Then (10y+x)+(10x+y)=121

⇒10y+x+10x+y=121

⇒11x+11y=121

⇒x+y=11

Thus the required linear equation is x + y = 11.

Answer 2

$\bf\green{QuestioN:-}$

The sum of a two-digit number and the number obtained  required by reversing its digits is 121. Find the number if it's  unit place digit is 5.​

$\bf\blue{AnsweR:-}$

The unit digit of the number = 5

The ten's place be x.

Then,

The number be 10x + 5.

The reciprocal be 50 + x.

According to the given condition,

$\bf\longrightarrow{(10x+5)+(50+x)=121}$

$\bf\longrightarrow{11x+55=121}$

$\bf\longrightarrow{11x=121-55}$

$\bf\longrightarrow{11x=66}$

$\bf\longrightarrow{x=\dfrac{66}{11}}$

$\bf\longrightarrow{x=6}$

Then the tenth place digit of the number is 6.

Unit place is already given as 5.

Hence,

The required number = 65.

Verification:-

According to the given condition,

$\bf\longrightarrow{(10x+5)+(50+x)=121}$

We find that x = 6.

On substituting the value of x,

$\bf\longrightarrow{(10\times6+5)+(50+6)=121}$

$\bf\longrightarrow{(65)+(56)=121}$

$\bf\longrightarrow{121=121}$

LHS = RHS

Hence Verified!!

Happy Learning!!