Question

old as
i). The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.​

Answer 1

❍ Let's say, that the unit place digit be x and ten's place digit be y respectively.

∴ Hence, the number is (10y + x)

A/Q,

• It is given that, the sum of the digits of a two – digit number is 9. Therefore,

➠ x + y = 9 ⠀⠀⠀⠀⠀⠀—eq ( I ).

Also,

• Nine times this Number is twice the number obtained by reversing the order of digits.

After Interchanging the digits, the number becomes : (10x + y).

Therefore,

★ 9 × (Number) = 2 × (Reversed no.) ★

⠀⠀⠀⠀⠀⠀

$:\implies\sf 9 \Big\{ 10y + x \Big\} = 2 \Big\{10x + y \Big\} \\\\\\:\implies\sf 90y + 9x = 20x + 2y \\\\\\:\implies\sf 20x + 2y - 90y - 9x = 0 \\\\\\:\implies\sf 11x - 88y = 0 \\\\\\:\implies\sf 11(x - 8y) = 0\\\\\\\:\implies\sf x - 8y = 0 \\\\\\:\implies\sf x = 8y \qquad\bigg\lgroup\frak Equation\;( \; I \;I\;)\bigg\rgroup$

⠀⠀⠀⠀⠀⠀

Now, from both equations, Substituting the value of x from equation ( I I ) in equation ( I ) :

⠀⠀⠀⠀⠀⠀

$:\implies\sf x + y = 9 \\\\\\:\implies\sf 8y + y = 9 \\\\\\:\implies\sf 9y = 9 \\\\\\:\implies\sf y = \cancel\dfrac{9}{9} \\\\\\:\implies\underline{\boxed{\pmb{\frak{\purple{y = 1}}}}}\:\bigstar$

⠀⠀⠀⠀⠀⠀

Now, Substituting the value of y in equation ( I ) :

⠀⠀⠀⠀⠀⠀

$:\implies\sf x + y = 9 \\\\\\:\implies\sf x + 1 = 9 \\\\\\:\implies\sf x = 9 - 1 \\\\\\:\implies\underline{\boxed{\pmb{\frak{\purple{x = 8}}}}}\;\bigstar$

⠀⠀⠀⠀⠀⠀

The number is (10y + x) :

➠ No. = 10 × (1) + 8

➠ No. = 10 + 8

➠ No. = 18

⠀⠀⠀⠀⠀⠀

$\therefore{\underline{\textsf{Hence,~the~required~number~is~\textbf{18}.}}}$

⠀⠀⠀⠀⠀⠀

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

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

• It is given that, the sum of the digits of a two – digit number is 9. So, let's verify the digits :

Therefore,

$\dashrightarrow\sf x + y = 9 \\\\\\\dashrightarrow\sf 8 + 1 = 9 \\\\\\\dashrightarrow\boxed{\frak{9 = 9}}$

⠀⠀⠀⠀⠀⠀

$\qquad\quad\therefore{\underline{\pmb{\sf{\blue{Hence, Verified..}}}}}$

Answer 2

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Correct \; Question:}}}}}}}$

⋆ The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Given \: that:}}}}}}}$

⋆ The sum of the digits of a two-digit number is 9.

⋆ Nine times this number is twice the number obtained by reversing the order of the digits

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{To \: find:}}}}}}}$

⋆ The original number

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Solution:}}}}}}}$

⋆ The original number = 18

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Assumptions:}}}}}}}$

⋆ Let a as the unit place.

⋆ Let b as the tens place.

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Full \; Solution:}}}}}}}$

⋆ The number be 10b+a because

${\sf{:\implies The \: unit \: place \: is \: a}}$

${\sf{:\implies The \: tens \: place \: is \: b}}$

~ Now as it is given that the sum of the digits of a two-digit number is 9. Henceforth,

${\sf{:\implies a+b \: = 9 \dots Eq. \: 1}}$

~ It is also given that nine times this number is twice the number obtained by reversing the order of the digits. Henceforth, after reversing it became the following

${\sf{:\implies (10a+b)}}$

~ Now according to the statement,

${\sf{:\implies 9 \times Number \: = 2 \times Reversed \: number}}$

${\sf{:\implies 9 \times (10b+a) \: = 2 \times (10a+b)}}$

${\sf{:\implies 90b + 9a \: = 20a + 2b}}$

${\sf{:\implies 90b - 2b \: = 20a - 9a}}$

${\sf{:\implies 88b \: = 11a}}$

${\sf{:\implies 11a - 88b \: = 0}}$

${\sf{:\implies 11(1)(a) - (8)(b) \: = 0}}$

${\sf{:\implies a-8b \: = 0}}$

${\sf{:\implies a \: = 0 + 8b}}$

${\sf{:\implies a \: = 8b \dots Eq. \: 2}}$

~ Now let's carry on, we have to take the help of ...Eq. 1 and ...Eq. 2 now Let us imply value of a now.

${\sf{:\implies a + b = 9}}$

${\sf{:\implies 8b + b = 9}}$

${\sf{:\implies 9b = 9}}$

${\sf{:\implies b = \dfrac{9}{9}}}$

${\sf{:\implies b = \cancel{\dfrac{9}{9}}}}$

${\sf{:\implies b = 1}}$

Henceforth, the value of b we get is 1

~ Now we have to take help of ...Eq. 1 Let us imply the value of b here.

${\sf{:\implies a + b = 9}}$

${\sf{:\implies a + 1 = 9}}$

${\sf{:\implies a = 9-1}}$

${\sf{:\implies a = 8}}$

Henceforth, the value of a we get is 8

~ Now at last let us put the value of a and b in the number that we get due to our assumptions.

${\sf{:\implies 10b+a}}$

${\sf{:\implies 10(1) + 8}}$

${\sf{:\implies 10 \times 1 + 8}}$

${\sf{:\implies 10 + 8}}$

${\sf{:\implies 18}}$

Henceforth, 18 is the number.