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 no.

Answer 1

let the no.s be x and y
x + y = 9 x = 9-y eq.1
ATQ
9(10x + y) = 2 ( 10y+x)
90x + 9y = 20y + 2x
88x = 11y
8x = y
8(9-y) = y
72 - 8y = y
72 = 9y
y = 8. eq.2
x + y = 9. From eq.1
x + 8 = 9
x = 1
So the no.s are 18 and 81

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{One's\;digit\;number\;be\;p}}$

Also,

$\textbf{\underline{Ten's\;digit\;number\;be\;n}}$

Situation,

p + n = 9

Also,

9(10n + p) = 2(10p + n)

Therefore,

p + n = 9

p = 9 - n

Hence,

9(10n + p) = 2(10p + n)

90n + 9p = 20p + 2n

88n = 11p

$\large{\boxed{\sf\:{Put\;the\;value\;of\;p}}}$

88n = 11(9 - n)

88n = 99 - 11n

99n = 99

n = 1

$\large{\boxed{\sf\:{Put\;the\;value\;of\;n}}}$

p = 9 - 1

p = 8

Therefore,

Number = 10n + p

10 + 8 = 18