Question

the digit of a two digit number differ by 3 if the digits are interchanged and the resulting number is added to the original number,we get 143. what can be the original number​

Answer 1

Solution :

Let, the units place digit is x and the tenth place digit is y with x > y

Then, the number is (10y + x)

If the digits be interchanged, the number becomes (10x + y)

By the given conditions,

x - y = 3 .....(i)

(10y + x) + (10x + y) = 143

or, 10y + x + 10x + y = 143

or, 11x + 11y = 143

or, x + y = 13 .....(ii)

Adding (i) and (ii), we get

x - y + x + y = 3 + 13

or, 2x = 16

or, x = 8

Putting x = 8 in (i), we get

8 - y = 3

or, y = 8 - 3

or, y = 5

Therefore, the original number can be

(10y + x) = {(10 * 5) + 8} = 50 + 8 = 58

Answer 2

Solution :-

Let the digit in ones place be x.

So, the digit in tens place be x + 3.

Original no. = 10(x + 3) + 1(x)

= 10x + 30 + x

= 11x + 30

New no. = 10(x) + 1(x + 3)

= 10x + x + 3

= 11x + 3

According to Question,

Original no. + New no. = 143

(11x + 30) + (11x + 3) = 143

⇒ 22x + 33 = 143

⇒ 22x = 143 - 33

⇒ 22x = 110

⇒ x = 110 / 22

⇒ x = 5

Required Numbers -

Original no. = 11x + 30

= 11(5) + 30

= 55 + 30

= 85

New no. = 58

Hence, the required number is 58.

Hope It Helps !!!