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(D) To minimize the sum $\frac{a}{b} + \frac{c}{d}$ where $a, b, c, d$ are distinct elements from $\{1, 2, 3, \ldots, 9\}$,we should choose the smalle

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$\frac{25}{72}$

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(D) To minimize the sum $\frac{a}{b} + \frac{c}{d}$ where $a, b, c, d$ are distinct elements from $\{1, 2, 3, \ldots, 9\}$,we should choose the smallest possible values for the numerators $a$ and $c$ and the largest possible values for the denominators $b$ and $d$.Let the set of values be $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.We choose $a=1, c=2$ (or vice versa) and $b=9, d=8$ (or vice versa).Then,the sum is $\frac{1}{9} + \frac{2}{8} = \frac{1}{9} + \frac{1}{4} = \frac{4+9}{36} = \frac{13}{36}$.Alternatively,if we choose $a=2, c=1$ and $b=9, d=8$,we get $\frac{2}{9} + \frac{1}{8} = \frac{16+9}{72} = \frac{25}{72}$.Comparing the values,$\frac{13}{36} = \frac{26}{72}$,which is greater than $\frac{25}{72}$.Thus,the minimum value is $\frac{25}{72}$.

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

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