Question

Let $W, X, Y,$ and $Z$ be four different digits selected from the set

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\dfrac{2}{17}$

$\dfrac{3}{17}$

$\dfrac{17}{72}$

$\dfrac{25}{72}$

$\dfrac{13}{36}$

Solution:

Small numerators and large denominators make small fractions, so use $1$ and $2$ as numerators and $8$ and $9$ as denominators.

Pairing the larger numerator with the larger denominator gives $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},$ which is smaller than $\dfrac19 + \dfrac28 = \dfrac{26}{72}.$ So the minimum sum is $\dfrac{25}{72}.$

Thus, the correct answer is D .

1994 AMC 8 Problem 20

Solution:

Problem 20 in Other Years