Question

the number obtained by interchanging the digit of a 2-digit number is 9 more than the orignal number. if the sum of the digits is 9, than find the orignal number.​

Answer 1

Given: The sum of the digits is 9. & The number obtained by Interchanging the digits of a two – digit number is 9 more than the Original number.

Need to find: The Original number?

⠀

❍ Let's consider that the two digits be x and y respectively.

Hence,

• Original number = (10x + y).

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$:\implies\sf x + y = 9$

$:\implies\sf y = 9-x\qquad\quad\bigg \lgroup \frak Eq^n \;( \: I \: ) \bigg\rgroup$

$\underline{\bigstar\:{\pmb{\mathcal{ACCORDING\; \: TO\; \: THE\; QUESTION\; :}}}}$

• The number obtained by Interchanging the digits of a two – digit number is 9 more than the Original number.

⠀

$\dashrightarrow\sf 10y + x = 10x + y + 9$

$\dashrightarrow\sf 9y = 9x + 9$

$\dashrightarrow\sf y = x + 1$

$\dashrightarrow\sf x + 1 = 9-x\qquad\quad\bigg \lgroup\sf From\;eq^n \;( \: I \: ) \bigg\rgroup$

$\dashrightarrow\sf 2x = 8$

$\dashrightarrow\sf x = \cancel\dfrac{8}{2}$

$\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{x =4}}}}}\:\bigstar$

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀⠀⠀⠀

$\dashrightarrow\sf x + y = 9$

$\dashrightarrow\sf 4 + y = 9$

$\dashrightarrow\sf y = 9 - 4$

$\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{y = 5}}}}}\;\bigstar$

✰ O r i g i n a l⠀N u m b e r :

⠀

$\twoheadrightarrow\sf Number = 10x + y$

$\twoheadrightarrow\sf Numebr = 10\Big(4\Big) + 5$

$\twoheadrightarrow\sf Number = 40 + 5$

$\qquad\twoheadrightarrow\boxed{\boxed{\pmb{\frak{\purple{45}}}}}\:\bigstar$

$\therefore{\underline{\textsf{Hence, the Original number is \textbf{45}.}}}$

Answer 2

$\large\underline{\sf{Given}}$

$\:$

» The number obtained by interchanging the digit of a 2 - digit number is 9 more than the original number.

» The sum of the digits of two digit number is 9.

$\:$

$\large\underline{\sf{To\:Find}}$

$\:$

» Original number?

$\:$

$\large\underline{\sf{Solution}}$

$\:$

• Let ten's digit of a number be m

• And one's digit of a number be n

• So, Original number = 10m + n

• And number obtained after interchanging digits = 10n + m

$\:$

$\underline{\sf{\bigstar\:According\:to\:the\:Question\::}}$

$\:$

$\sf \longrightarrow \: Sum\:of\:digits\:of\:two\:digit\:number\:=\:9$

$\\ \sf \longrightarrow \: m + n = 9$

$\\ \sf \longrightarrow \: \pink {m = 9 - n}\qquad\big\lgroup 1 \big\rgroup$

$\:$

Also,

$\:$

$\sf \longrightarrow \: Interchanged\:number = \big(Original\:number\big)\:+\:9$

$\\ \sf \longrightarrow \: 10n + m = \big(10m + n\big) + 9$

$\\ \sf \longrightarrow \: 10n + m = 10m + n + 9$

$\\ \sf \longrightarrow \: 10n - n = 10m - m + 9$

$\\ \sf \longrightarrow \: 9n = 9m + 9$

$\\ \sf \longrightarrow \: 9n = 9\big(m + 1\big)$

$\\ \sf \longrightarrow \: \dfrac{\cancel{9}n}{\cancel{9}} = m + 1$

$\\ \sf \longrightarrow \: n = m + 1$

$\\ \sf \longrightarrow \: \pink{m = n - 1} \qquad\big\lgroup 2 \big\rgroup$

$\:$

From [1] and [2], we get,

$\:$

$\sf \longrightarrow \: 9 - n = n - 1$

$\\ \sf \longrightarrow \: 9 + 1 = n + n$

$\\ \sf \longrightarrow \: 2n = 10$

$\\ \sf \longrightarrow \: n = {\cancel{\dfrac{10}{2}}}$

$\\ \sf \longrightarrow \: \pink{ n = 5}$

$\:$

Putting value of 'n' in [1],

$\:$

$\sf \longrightarrow \: m = 9 - 5$

$\\ \sf \longrightarrow \:\pink{ m = 4 }$

$\:$

Finding original number,

$\:$

$\sf \longrightarrow \: Original\:number = 10m + n$

$\:$

Putting values of 'm' and 'n',

$\:$

$\sf \longrightarrow \: Original\:number = 10\big(4\big) + 5$

$\\ \sf \longrightarrow \: Original\:number = \big(10\:\times\:4\big) + 5$

$\\ \sf \longrightarrow \: Original\:number = 40 + 5$

$\\ \sf \longrightarrow \:\pink{ Original\:number = 45}$

$\:$

$\therefore\:{\underline{\sf{Original\:number\:is\:{\textsf{\textbf{45}}}}}}$

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