Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 121 and the two digits differ by 3. Find the number

Answer 1

let the digit at ten's place be x. it's value= 10x
digit at one's place(and it's value)= x-3
the number= (10x) + (x-3)
number after reversing it= 10(x-3) + x

A.T.Q          (10x) + (x-3) + 10(x-3) + x = 121
               ⇒ 10x + x - 3 + 10x - 30 + x = 121
               ⇒  22x = 121+33
               ⇒ x = 154/22
                 ∴ x= 7

so, the digit at ten's place = 7
digit at one's place = 7-3 = 4
the number = (7×10) + 4
                   = 74

so, the required number is 74 or 37.

Answer 2

Answer:

Step-by-step explanation:

Given :-

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

The two digits differ by 3.

To Find :-

The Number

Solution :-

Let x be the ones digit.

y be the tens digit.

Two digit number before reversing = 10y + x  

Two digit number after reversing = 10x + y  

According to the question

1st Equation

⇒ (10y + x) + (10x + y) = 121  

⇒ 11x + 11y = 121  

⇒ x + y = 11 …….(i)  

2nd Equation

Since the digits differ by 3, so  

x – y = 3 ……….(ii)  

Adding Eq (i) and (ii),

⇒ 2x = 14  

⇒ x = 7  

Putting x value in Eq (i)

⇒ x + y = 11

⇒ 7 + y = 11  

⇒ y = 4  

Changing the role of x and y,

x = 4 and y = 7  

Hence, the two-digit number is 74 or 47.