11. The digits of a two-digit number differ by 3. If digits are interchanged and theresulting number is added to the original number, we get 121. Find the originalnumber. About the author Jasmine
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 Reply
❍ Let’s say, that the one’s place digit be y and ten’s place digit be x respectively. Original number = (10x + y). ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ 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 ) ⠀⠀⠀ [tex]\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}[/tex] ⠀⠀⠀ If digits are interchanged and the resulting number is added to the original number, we get 121. ⠀⠀⠀ Therefore, ⠀⠀⠀ [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] ⠀⠀⠀ ⠀⠀⠀[tex]\underline{\bf{\dag} \:\mathfrak{Putting\; value \;of\; y \;in\;eq^{n}\;(1)\: :}}[/tex]⠀⠀⠀⠀ ⠀⠀⠀ [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) ✰ ⠀⠀⠀ ⇥ No. = 10x + y ⇥ No. = 10(7) + 4 ⇥ No. = 70 + 4 ⇥ No. = 74 ⠀⠀⠀ ∴ 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 : As it is given that, the digits of a two– digit number is differ by 3. Therefore, [tex]:\implies\sf x – y = 3 \\\\\\:\implies\sf 7 – 4 = 3 \\\\\\:\implies\underline{\boxed{\frak{ 3 = 3}}}[/tex] ⠀⠀⠀ [tex]\qquad\quad\therefore{\pink{\underline{\textsf{\textbf{Hence, Verified!}}}}}[/tex] Reply
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 )
⠀⠀⠀
[tex]\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}[/tex]
⠀⠀⠀
⠀⠀⠀
Therefore,
⠀⠀⠀
[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]
⠀⠀⠀
⠀⠀⠀[tex]\underline{\bf{\dag} \:\mathfrak{Putting\; value \;of\; y \;in\;eq^{n}\;(1)\: :}}[/tex]⠀⠀⠀⠀
⠀⠀⠀
[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) ✰
⠀⠀⠀
⇥ No. = 10x + y
⇥ No. = 10(7) + 4
⇥ No. = 70 + 4
⇥ No. = 74
⠀⠀⠀
∴ 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]
⠀⠀⠀
[tex]\qquad\quad\therefore{\pink{\underline{\textsf{\textbf{Hence, Verified!}}}}}[/tex]