Question

The sum of two numbers is $18$. The sum of their reciprocals is $1/4$. The smaller number is

Answer 1

$\Rightarrow$ Let one number be $x$, then other number will be $18-x$.

According to the question,

$\Rightarrow$ $\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}$

$\Rightarrow$ $\frac{18-x+x}{18x-x^2}=\frac{1}{4}$

$\Rightarrow$ $\frac{18}{18x-x^2}=\frac{1}{4}$

$\Rightarrow$ $72=18x-x^2$

$\Rightarrow$ $x^2-18x+72=0$

$\Rightarrow$ $x^2-12x-6x+72=0$

$\Rightarrow$ $x(x-12)-6(x-12)=0$

$\Rightarrow$ $(x-12)(x-6)=0$

$\Rightarrow$ $x=12$ and $x=6$

$\Rightarrow$ Let $x=6$ then, $18-x=12$

$\therefore$ Two number are $6$ and $12$

$\Rightarrow$ Smallest number is $6$.