Question

A two digit number and the number with digits interchanged add up to
143. In the given number the digit in unit’s place is 3 more than the
digit in the ten’s place. Find the original number.

Let the digit in unit’s place is x

and that in the ten’s place is y

∴ the number = ☐ y + x

The number obtained by interchanging the digits is ☐ x + y

According to first condition two digit number + the number obtained by
interchanging the digits = 143

∴ 10y+x + ☐ =143

∴ ☐ x + ☐y =143

x+y = ☐.......... (I)

From the second condition,
digit in unit’s place = digit in the ten’s place + 3

∴ x =☐ +3
∴ x-y =3 ....... (II)

Adding equations (I) and (II)

2x = ☐
x=8
Putting this value of x in equation (I)
x + y =13
8 + ☐ =13
∴ y = ☐
The original number is 10 y + x
= ☐ +8
= 58

Answer 1

Let the digit in unit’s place is x

and that in the ten’s place is y

∴ the number = $\bold{\boxed{10}}$ y + x

The number obtained by interchanging the digits is $\bold{\boxed{10}}$ x + y

According to first condition two digit number + the number obtained by
interchanging the digits = 143

∴ 10y+x + $\bold{\boxed{10x + y}}$ =143

∴ $\bold{\boxed{11}}$x + $\bold{\boxed{11}}$y =143

x + y = $\bold{\boxed{13}}$.......... ( 1 )

From the second condition,
digit in unit’s place = digit in the ten’s place + 3

∴ x =$\bold{\boxed{y}}$ +3
∴ x - y = 3 ....... ( 2 )

Adding equations (I) and (II)
x - y = 13
x + y = 3
--------------
2x = $\bold{\boxed{16}}$
x=8
Putting this value of x in equation (I)
x + y =13
8 + $\bold{\boxed{y}}$ =13
∴ y = $\bold{\boxed{13 - 8 = 5}}$

The original number is 10 y + x
= $\bold{\boxed{10*5}}$ +8
= 58

Answer 2

✴✴$\bf{Hey \: friends!!}$✴✴
---------------------------------------------------------

✴✴$\underline{Here \: is \: your \: answer↓}$
⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇

▶⏩

$\bf{Let \: the \: unit's \: digit \: be \: x.}$
$\bf{and \: the \: ten's \: digit \: be \: y.}$

$\boxed{The \: real \: number \: = x + 10y.}$
$\boxed{and \: the \: interchanged \: number \: = 10x + y}$
$\bf{A/Q}$
$\bf{In \: 1st \: case:}$
$\bf{Real \: number + interchanged \: number = 143.}$
↪➡ x + 10y + 10x + y = 143.

↪➡ 11x + 11y = 143.

↪➡ 11( x + y ) = 143.

↪➡ x + y = 143/11.

$\boxed{x + y = 13..................(1).}$

$\bf{In \: 2nd \: case \: : }$
$\bf{unit's \: digit = ten's \: digit + 3.}$
↪➡ x = y + 3

$\boxed{x - y = 3...................(2)}$
▶⏩ Substract in equation (1) and (2).

x + y = 13.
x - y = 3.
- + -
_________
2y = 10.

↪➡ y = 10/2.

$\underline{y = 5.}$
▶⏩ Put the value of ‘y’ in equation (1).

↪➡ x + 5 = 13.

↪➡ x = 13 - 5.

$\underline{x = 8.}$
✔✔Hence, the real number =

= x + 10y

= 8 + 10× 5.

$\boxed{= 58.}$

✴✴$\boxed{THANKS}$✴✴

☺☺☺$\underline{Hope \: it \: is \: helpful \: for \: you}$✌✌✌.