Question

6. The sum of the digits of a two-digit number is 11. If we interchange
the digits then the new number formed is 45 less than the original. Find
the original number.​

Answer 1

AnswEr:-

Two digit number = 83

$\rule{250}{1}$

Let the digit at unit's place be m & digit at ten's place be n

⇒ Original number = m + 10n

Case 1:-

⇒ m + n = 11 [Eq.1]

Case 2:-

★ Interchanged number :-

⇒ New number = n + 10m

According to question:-

⇒ Original number - New number = 45

⇒ m + 10n - (n + 10m) = 45

⇒ m + 10n - n - 10m = 45

⇒ 9n - 9m = 45

⇒ 9(n - m) = 45

⇒ n - m = 45/9

⇒ n - m = 5 [Eq.2]

Now adding both equations:-

$\qquad\sf{n \: + \: \cancel{m} = 11}$

$\qquad\sf{n \: - \: \cancel{m} = 5}$

$\quad\sf{\: (+) \: \: (-) \: \: (+)}$

$\: \: \rule{100}{1}$

$\qquad\sf{\: 2n = \:\: 16}$

$\twoheadrightarrow\sf {\: \: \: \: \: n = 16/2}$

$\twoheadrightarrow\sf{\: \: \: \: \: n =\large {\boxed{\sf\green{8}}}}$

So, digit at ten's place = 8

Put this value in (Eq.2)

⇒ 8 - m = 5

⇒ m = 8 - 5

⇒ m = 3

So, digit at unit's place = 3

Now number:-

⇒ Original number = 3 + 10(8)

⇒ Original number = 3 + 80

⇒ Original number = 83

Therefore,

$\therefore{\underline{\boxed{\textsf{Original number = {\textbf{83 }}}}}}$

Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

The sum of the digits of a two - digits number is 11. If we interchange the digits then the new number formed is 45 less than the original number.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The original number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the digit of ten's pace be r

Let the digit of one's pace be m

So;

$\sf{\underline{The\:Original\:number=\pink{10r+m}}}\\\sf{\underline{The\:Reversed\:number=\pink{10m+r}}}$

A/q

$\mapsto\bf{r+m=11.........................(1)}$

&

$\mapsto\sf{10r+m-(10m+r)=45}\\\\\mapsto\sf{10r+m-10m-r=45}\\\\\mapsto\sf{10r-r+m-10m=45}\\\\\mapsto\sf{9r-9m=45}\\\\\mapsto\sf{9(r-m)=45}\\\\\mapsto\sf{r-m=\cancel{\dfrac{45}{9} }}\\\\\mapsto\sf{r-m=5}\\\\\mapsto\bf{r=5+m..........................(2)}$

Putting the value of r in equation (1),we get;

$\mapsto\sf{5+m+m=11}\\\\\mapsto\sf{5+2m=11}\\\\\mapsto\sf{2m=11-5}\\\\\mapsto\sf{2m=6}\\\\\mapsto\sf{m=\cancel{\dfrac{6}{2} }}\\\\\mapsto\sf{\orange{m=3}}$

Putting the value of m in equation (2),we get;

$\mapsto\sf{r=5+3}\\\\\mapsto\sf{\orange{r=8}}$

Thus;

$\underbrace{\sf{The\:original\:number\:is\:10r+m=10(8)+3=80+3=83}}}}$.