Question

the sum of two digits number and the number obtained by reversing the order of its digits is 143. the digits at ten's 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

Answer 2

✨welcome to the concept of linear equations

Let the number at the units place be X and the digit at the tens place be y.

then ,the number ,

is 10y+X

after interchang the digits the number becomes 10x+y.

given :

the sum of two digits number and the number obtained by reversing the order of its digits is 143.

then ,

10y+X+ 10x+y = 143

11x+11y= 143

x+y = 13

or. y+X= 13 .....(1)

also ,given :

the digits at ten's place is greater than digit of units place by 3 .

then ,

y = X+3

then,

y-x= 3.....(2)

✨ now solve equations (1) and (2)

y+X= 13

y-x= 3

add these two equations

then

2y= 16

y= 16/2= 8

put y = 8 in equation (2)

y-x=3

X=y-3

X= 8-3 = 5

then orignal no is 10y+X= 10×8+5

= 85

I hope it help you ❤️

in case of doubt ask to me

my dear friend ❤️