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

Answer: (D)

$\frac{25}{72}$

To minimize the value of $\frac{a}{b}+\frac{c}{d}$, $b$ and $d$ must highest numbers chosen from the set and $a$ and $c$ must be the lowest numbers of given set.

Therefore, $b=9,d=8,a=2$ and $c=1$.

Thus, the minimum value for $\frac{a}{b}+\frac{c}{d}$ $=\frac{2}{9}+\frac{1}{8}=\frac{25}{72}$