11. The digits of a two-digit number differ by 3. If digits are interchanged and the
resulting number is added to the original number, we get 121. Find the original
number.
11. The digits of a two-digit number differ by 3. If digits are interchanged and the
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Answer:
Given :–
The digits of a two-digit number differ by 3. If digits are interchanged and the
resulting number is added to the original number, we get 121
To Find :–
Original number
Solution :–
Let us assume that the digit at unit place is x.
Now
The tens digit is will be 3 more than it. So, the digit becomes x + 3
According to the question
[tex] \sf \: Number = \bigg( 10(x + 3) \bigg) + x[/tex]
[tex] \sf \: Number = 10x + 30 + x[/tex]
[tex] \sf \: Number = (10x + x) + 30[/tex]
[tex] \sf \: Number = 11x + 30[/tex]
When the number interchange
[tex] \sf \: Interchanged \: Number = 10 + (x + 3)[/tex]
[tex] \sf \: Interchanged \: Number = (10x + x) + 3[/tex]
[tex] \sf \: Interchanged \: Number = 11x + 3[/tex]
Adding both the equation
[tex] \sf \: (11x + 30) + (11x + 3) = 121[/tex]
[tex] \sf \: (11x + 11x) + (30 + 3) = 121[/tex]
[tex] \sf \: 22x + 33 = 121[/tex]
[tex] \sf \: 22x = 121 – 33[/tex]
[tex] \sf \: 22x = 88[/tex]
[tex] \sf \: x = \cancel \dfrac{88}{22} [/tex]
[tex] \sf \: x = 4[/tex]
Now
The tens digit = 4 + 3 = 7
Hence
The number is 74
❍ Let’s say, that the one’s place digit be y and ten’s place digit be x respectively.
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀
Given that,
Then,
➟ x – y = 3
➟ x = y + 3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( I )
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[tex]\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}[/tex]
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Therefore,
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[tex]\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 [/tex]
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⠀⠀⠀[tex]\underline{\bf{\dag} \:\mathfrak{Putting\; value \;of\; y \;in\;eq^{n}\;(1)\: :}}[/tex]⠀⠀⠀⠀
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[tex]\dashrightarrow\sf x = y + 3\\\\\\\dashrightarrow\sf x = 4 + 3\\\\\\\dashrightarrow\boxed{\frak{\pink{\pmb{\purple{x = 7}}}}}\;\bigstar[/tex]
✰ 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.
[tex]\rule{250px}{.3ex}[/tex]
V E R I F I C A T I O N :
Therefore,
[tex]:\implies\sf x – y = 3 \\\\\\:\implies\sf 7 – 4 = 3 \\\\\\:\implies\underline{\boxed{\frak{ 3 = 3}}}[/tex]
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[tex]\qquad\quad\therefore{\pink{\underline{\textsf{\textbf{Hence, Verified!}}}}}[/tex]