Question

The digit of two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number.
​

Answer 1

Answer:

Step-by-step explanation:

let the digit in unit place is x

and the digit at tens place is y

therefore the number so obtained is x+10y

now interchanging the digits means x is in tens place and y is in units place

number so obtained is 10x+y

adding these two numbers we get

x+10y+10x+y = 121

11x+11y=121

x+y=11

now difference between the two digits of the number is 3

therefore

y-x=3

adding two equations we get,

2y=14

y=7

x+7=11

x=4

therefore the number ie 74 or 47

Answer 2

$\huge\bf\underline{\red{A}\green{N}\orange{S}\pink{W}\purple{ER}}$

Given :-

• The digits of a two digits number differ by 3. If the digits are interchanged and the resulting number is added to the original number we get 121.

To Find :-

• What is the number.

Solution :-

Let, the one's place number be x

And, the ten's place number will be x + 3

So, the original number will be,

↦ 10(x - 3) + x

↦ 10x - 30 + x

➤ 11x - 30

When the number is interchanged then the number will be 10x + x - 3 = 11x - 3

According to the question,

⇒ 11x - 3 + 11x - 30 = 121

⇒ 11x + 11x - 3 - 30 = 121

⇒ 22x - 33 = 121

⇒ 22x = 121 + 33

⇒ 22x = 154

⇒ x = 154 ÷ 22

➠ x = 7

Hence, the required number will be,

↪ 11x - 30

↪ 11(7) - 30

↪ 77 - 30

➦ 47

∴ The original number will be 47 .