Question

The sum 2 digit number obtained by reversing the order of its digit is 143 . The digit at tens place is greater than digit of units place by 3 then find the original number​

Answer 1

Answer:

Let, Digit at ones place=x

therefore,Digit at tens place=3x

Original Number= 10.3x +1.x

=30x+x

=31x

New number= 10.x+ 1.3x

=10x+3x

=13x

A/c to Question:

31x +13x = 143

44x = 143

x= 143/44

x=3.25

Hence,Digit at ones place =3.25

Digit at tens place=3(3.25)

=9.75

Rounding the digits=103

Step-by-step explanation:

please make brainlist answer and please follow me