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.
ʙy ᴇʟɪᴍɪɴᴀᴛɪᴏɴ ᴍᴇᴛʜᴏᴅ..​

Answer 1

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$\mathfrak{let \: the \: units \: place \: digit \: be \: x} \\ \ \: \mathfrak{tens \: place \: digit \: be \: y} \\ \: \mathfrak{ \: therefore \: the \: number \: formed \: = 10y + x} \\ \\\mathfrak{ \: number \: formed \: after \: reversing \: = 10x + y} \\ \\ \\\mathfrak{according \: to \: question} \\ x + y = 9 - - - (i) \\ \implies \: 9(10y + x) = 2(10x + y) \\ \implies90y + 9x = 20x - 2y \\ \implies88y - 11x = 0 \\ \implies - x + 8y = 0 - - - (ii) \\ \\ (i) + (ii)\implies \: x + y - x + 8y = 9 \\\implies \: y + 8y = 9 \\ \implies \: 9y = 9 \\\implies \: y = 1 - - - (iii) \\ \\ \mathfrak{putting \: the \: value \: of \: y \: in \: equation \: (i)} \\ x = 8 \\ \\\mathfrak{ \: therefore \: the \: number = 10 \times 1 + 8 = 18}$

Answer 2

Answer:

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