Question

(ii) 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

Given:

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.

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☯ Let the unit digit and tens digits of the number be x and y respectively.

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Therefore,

• Number = 10y + x

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After reversing,

• The number become = 10x + y

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${\underline{\sf{\bigstar\; According\;to\; Question\;:}}}$

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

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$:\implies\sf x + y = 9\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

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

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$:\implies\sf 9(10y + x) = 2(10x + y)$

$:\implies\sf 90y + 9x = 20x + 2y$

$:\implies\sf 90y - 2y = 20x - 9x$

$:\implies\sf 88y = 11x$

$:\implies\sf 11x - 88y = 0$

$:\implies\sf 11(x - 8y) = 0$

$:\implies\sf x - 8y = 0\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

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★ Now, Adding equation (i) and (ii),

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$:\implies\sf 9y = 9$

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

$:\implies{\boxed{\frak{\pink{y = 1}}}}\;\bigstar$

★Now, Putting value of y in equation (i),

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$:\implies\sf x + 1 = 9$

$:\implies\sf x = 9 - 1$

$:\implies{\boxed{\frak{\purple{x = 8}}}}\;\bigstar$

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Therefore,

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The number is,

• 10y + x = 10 × 1 + 8 = 10 + 8 = 18

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$\therefore\;{\underline{\sf{Hence,\;the\; required\;number\;is\; \bf{18}.}}}$