6. The sum of the digits of a two-digit number is 11. If we interchang

1.	6. The sum of the digits of a two-digit number is 11. If we interchangethe digits then the new number formed is 45 less than the original. Findthe original number.
Answer» 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:- $\qquad\sf{n \: + \: \cancel{m} = 11}$ $\qquad\sf{n \: - \: \cancel{m} = 5}$ $\quad\sf{\: (+) \: \: (-) \: \: (+)}$ $\: \: \rule{100}{1}$ $\qquad\sf{\: 2n = \:\: 16}$ $\twoheadrightarrow\sf {\: \: \: \: \: n = 16/2}$ $\twoheadrightarrow\sf{\: \: \: \: \: n =\large {\boxed{\sf\green{<klux>8</klux>}}}}$ 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, $\therefore{\underline{\boxed{\textsf{Original number = {\textbf{83 }}}}}}$

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

Answer»

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:-

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

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

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

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

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

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

$\twoheadrightarrow\sf{\: \: \: \: \: n =\large {\boxed{\sf\green{<klux>8</klux>}}}}$

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,

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