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

ToFind:–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 beyand ten’s place digit bexrespectively.(10x + y).⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀Given that,

3.Then,

➟ x – y = 3

➟

x = y + 3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( I )⠀⠀⠀

[tex]\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}[/tex]

⠀⠀⠀

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. is74.[tex]\rule{250px}{.3ex}[/tex]

V E R I F I C A T I O N :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]