Question

The sum of a two-digit number and the number obtained by reversing its digits is 121. Find the number if it’s unit place digit is 5.
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Answer 1

To Find :

The number considered in the question if the the sum of itself and its reverse is 121.

Let's Assume :

We are given that , Unit digit of the number = 5 .

So, let's assume that The number at tens digit = x

So, we get the number as = 10x+5

Solution :

As Given, the sum of the number itself and its reverse is 121 .

We can equate it as :

(10x+5 )+(50+x )

Now, using this equation, we will find the value of x .

$\tt{ \implies 10x + 5 + 50 + x = 121}$

$\tt{ \implies 11x + 55 = 121}$

$\tt{ \implies 11x = 121 - 55}$

$\tt{ \implies 11x = 66}$

$\large \tt{ \implies x = \frac{66}{11} }$

$\red{ \underline{ \boxed{\tt{ \implies x = 6}}}}$

Hence, The number = 10x+5

= 60+5

= 65.

Hence, the number considered here = 65.

Answer 2

Answer:

Giv€n Question :-♀️

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

__________________________

To Find :♀️

The number considered in the question if the the sum of itself and its reverse is 121.

Let's Assume :♂️

We are given that , Unit digit of the number = 5 .

So, let's assume that The number at tens digit = x

So, we get the number as = 10x+5

Solution :⤵️

As Given, the sum of the number itself and its reverse is 121 .

We can equate it as :

(10x+5 )+(50+x )

Now, using this equation, we will find the value of x .

$\rm{ \implies 10x + 5 + 50 + x = 121}$

$\rm{ \implies 11x + 55 = 121}$

$\rm{ \implies 11x = 121 - 55}$

$\rm{ \implies 11x = 66}$

$\large \rm{ \implies x = \frac{66}{11} }$

$\blue{ \underline{ \boxed{\sf{ \implies x = 6}}}}$

Hence, The number = 10x+5

= 60+5

= 65.

• Hence, the number considered here = 65.

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