Question

A certain number of two digits is four times the sum of the digits if 8 be added to the number and the digits are reversed the number is.... ​

Answer 1

${\boxed{\underline{\sf Answer \:is \:24}}}$

$\sf Let\:the\: tens\: digit \:number\: be\: x$

$\sf Let\: the\: unit \:digit\: number\: be\: y$

$\sf\bold{\underline{Case\: 1}}$

$\sf\implies{10x+y=4(x+y)}$

$\sf\longrightarrow{y=2x------------(i)}$

$\sf\bold{\underline{Case \:2}}$

$\sf\implies{10x+y+8=10y+x}$

$\sf\longrightarrow{ x-y=\frac{-8}{9}------------(ii)}$

$\sf\underline{Putting\:the\:value\:of\:y\:from\:(i)\:in\:(ii)}$

$\sf x-2x=\frac{-8}{9}$

$\sf-x=\frac{-8}{9}$

$\sf\implies x=\frac{8}{9}$

$\sf y=2x$

$\sf y=2\times {\frac{8}{9}}$

$\sf\implies y=\frac{16}{9}$

The digits are in fraction so forming a number using these digits is not possible so your question is wrong. but in case 2 if we add 18 instead of 8 then we can get a genuine answer. So, let's solve it by adding 18 in Case 2.

$\sf\bold{\underline{Case \:2}}$

$\sf\implies{10x+y+18=10y+x}$

$\sf\longrightarrow{ x-y=-2------------(ii)}$

$\sf\underline{Putting\:the\:value\:of\:y\:from\:(i)\:in\:(ii)}$

$\sf x-2x=-2 \\\\ -x=-2\\\\\implies x=2 \\\\y=2x\\\\ y=2\times2\\\\\implies y=4$

$\sf\bold{\underline{Hence,\:the\:number \:is\:24}}$