If each of the four numbers $3,4,6$, and 7 replaces a $\square$, what is the largest possible sum of the fractions shown?
(A) $\frac{19}{12}$
(D) $\frac{15}{4}$
(B) $\frac{13}{7}$
(E) $\frac{23}{6}$
(C) $\frac{5}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-058.jpg?height=100&width=130&top_left_y=568&top_left_x=1366)
The value of any positive fraction is increased by increasing the numerator and/or decreasing the denominator.

Thus, to obtain the largest possible sum, we choose 6 and 7 as the numerators, and 3 and 4 as the denominators.

We then calculate:

$$
\frac{7}{3}+\frac{6}{4}=\frac{28}{12}+\frac{18}{12}=\frac{46}{12}=\frac{23}{6}
$$

We recognize that $\frac{23}{6}$ is the largest of the 5 possible answers, and thus is the correct response. (This means that we do not need to try $\frac{7}{4}+\frac{6}{3}$.)

ANSWER: (E)