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Answers
Answer:
Let's say, that the one's place digit be y and ten's place digit be x respectively.
Original number = (10x + y).
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Given that,
As per given condition, the digits of a two – digit number is differ by 3.
Then,
➟ x – y = 3
➟ x = y + 3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( I )
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\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}
★According to the Question:
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If digits are interchanged and the resulting number is added to the original number, we get 121.
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Therefore,
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\begin{gathered}\dashrightarrow\sf 10x + y + 10y + x = 121 \\\\\\\dashrightarrow\sf 11x + 11y = 121 \\\\\\\dashrightarrow\sf x + y = 11\\\\\\\dashrightarrow\sf y + 3 + y = 11\qquad\quad\bigg\lgroup\sf From\;eq^{n}\;1\bigg\rgroup\\\\\\\dashrightarrow\sf 2y + 3 = 11\\\\\\\dashrightarrow\sf 2y = 11 - 3 \\\\\\\dashrightarrow\sf 2y = 8\\\\\\\dashrightarrow\sf y = \cancel\dfrac{8}{2}\\\\\\\dashrightarrow\underline{\boxed{\frak{\pink{\pmb{\purple{y = 4}}}}}}\;\bigstar \end{gathered}
⇢10x+y+10y+x=121
⇢11x+11y=121
⇢x+y=11
⇢y+3+y=11
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Fromeq
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1
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⇢2y+3=11
⇢2y=11−3
⇢2y=8
⇢y=
2
8
⇢
y=4
y=4
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⠀⠀⠀\underline{\bf{\dag} \:\mathfrak{Putting\; value \;of\; y \;in\;eq^{n}\;(1)\: :}}
†Puttingvalueofyineq
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\begin{gathered}\dashrightarrow\sf x = y + 3\\\\\\\dashrightarrow\sf x = 4 + 3\\\\\\\dashrightarrow\boxed{\frak{\pink{\pmb{\purple{x = 7}}}}}\;\bigstar\end{gathered}
⇢x=y+3
⇢x=4+3
⇢
x=7
x=7
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✰ ORIGINAL NO. = (10x + y) ✰
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⇥ No. = 10x + y
⇥ No. = 10(7) + 4
⇥ No. = 70 + 4
⇥ No. = 74
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∴ Hence, the required two – digit no. is 74.
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V E R I F I C A T I O N :
As it is given that, the digits of a two– digit number is differ by 3.
Therefore,
\begin{gathered}:\implies\sf x - y = 3 \\\\\\:\implies\sf 7 - 4 = 3 \\\\\\:\implies\underline{\boxed{\frak{ 3 = 3}}}\end{gathered}
:⟹x−y=3
:⟹7−4=3
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3=3
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\qquad\quad\therefore{\pink{\underline{\textsf{\textbf{Hence, Verified!}}}}}∴