Question

A number cosits of two digit of which tens digit exceeds the unit digit by 3. The number is ten time sum of it's digit find number

Answer 1

Let the ten's digit and the one's digit be x and y respectively.

Given

Ten's digit exceeds the unit digit by 3.

$x=y+3\\\implies x-y=3--(1)$

The number is ten times the sum of it's digits

$10x+y=10(x+y)\\\implies 10x+y=10x+10y\\\implies y=10y\\\implies y -10y=0\\\implies -9y=0\\\implies y = 0$

Substituting, the value of y in the first equation, we get,

$\implies x = 3$

$\boxed{\boxed{\bold{Therefore. \ the \ required \ number \ is \ 30}}}}}$

                                               

Answer 2

Answer:

Let the ten's digit & unit's digit be x and y.

⋆ Number = 10x + y

☯ According to First Statement :

⇢ Number = 10 × Sum of Digits

⇢ 10x + y = 10(x + y)

⇢ 10x + y = 10x + 10y

⇢ 10x – 10x = 10y – y

⇢ 0 = 9y

• Dividing both term by 9

⇢ y = 0

━━━━━━━━━━━━━━━

☯ According to Second Statement :

⇾ Ten's Digit = 3 + Unit's Digit

⇾ x = 3 + y

⇾ x = 3 + 0

⇾ x = 3

∴ Hence, Required Number will be 30.