Question

If $a,b,c,d$ are four distinct numbers chosen from the set $\{1,2,3....,9\},$ then the minimum value of $\frac{a}{b}+\frac{c}{d}$ is

• A. $\frac{3}{8}$
• B. $\frac{1}{3}$
• C. $\frac{13}{36}$
• D. $\frac{25}{72}$

Answer 1

The correct option is D $\frac{25}{72}$

For minimum of $\frac{a}{b}+\frac{c}{d},$
$a$ and $c$ should be smallest, and
$b$ and $d$ should be largest.
So, minimum value is $\frac{1}{9}+\frac{2}{8}=\frac{26}{72}$ or $\frac{1}{8}+\frac{2}{9}=\frac{25}{72}$
Hence, the minimum value is $\frac{25}{72}$