Question

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.​

Answer 1

❍ 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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$\underline{\bigstar\:\boldsymbol{According~to~ the~Question :}}$

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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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$\dashrightarrow\sf 10x + y + 10y + x = 121 \\\\\\\dashrightarrow\sf 11x + 11y = 121 \\\\\\\dashrightarrow\sf x + y = 11\\\\\\\dashrightarrow\sf y + 3 + y = 11\qquad\quad\bigg\lgroup\sf From\;eq^{n}\;1\bigg\rgroup\\\\\\\dashrightarrow\sf 2y + 3 = 11\\\\\\\dashrightarrow\sf 2y = 11 - 3 \\\\\\\dashrightarrow\sf 2y = 8\\\\\\\dashrightarrow\sf y = \cancel\dfrac{8}{2}\\\\\\\dashrightarrow\underline{\boxed{\frak{\pink{\pmb{\purple{y = 4}}}}}}\;\bigstar$

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⠀⠀⠀$\underline{\bf{\dag} \:\mathfrak{Putting\; value \;of\; y \;in\;eq^{n}\;(1)\: :}}$⠀⠀⠀⠀

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$\dashrightarrow\sf x = y + 3\\\\\\\dashrightarrow\sf x = 4 + 3\\\\\\\dashrightarrow\boxed{\frak{\pink{\pmb{\purple{x = 7}}}}}\;\bigstar$

✰ 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.

$\rule{250px}{.3ex}$

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,

$:\implies\sf x - y = 3 \\\\\\:\implies\sf 7 - 4 = 3 \\\\\\:\implies\underline{\boxed{\frak{ 3 = 3}}}$

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$\qquad\quad\therefore{\pink{\underline{\textsf{\textbf{Hence, Verified!}}}}}$

Answer 2

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

$\sf \: Number = \bigg( 10(x + 3) \bigg) + x$

$\sf \: Number = 10x + 30 + x$

$\sf \: Number = (10x + x) + 30$

$\sf \: Number = 11x + 30$

When the number interchange

$\sf \: Interchanged \: Number = 10 + (x + 3)$

$\sf \: Interchanged \: Number = (10x + x) + 3$

$\sf \: Interchanged \: Number = 11x + 3$

Adding both the equation

$\sf \: (11x + 30) + (11x + 3) = 121$

$\sf \: (11x + 11x) + (30 + 3) = 121$

$\sf \: 22x + 33 = 121$

$\sf \: 22x = 121 - 33$

$\sf \: 22x = 88$

$\sf \: x = \cancel \dfrac{88}{22}$

$\sf \: x = 4$

Now

The tens digit = 4 + 3 = 7

Hence

The number is 74