Ginger walks at $4 \mathrm{~km} / \mathrm{h}$ and runs at $6 \mathrm{~km} / \mathrm{h}$. She saves $3 \frac{3}{4}$ minutes by running instead of walking from her home to her school. What is the distance, in kilometres, from her home to her school?
(A) $7 \frac{1}{2}$
(B) $3 \frac{3}{4}$
(C) $1 \frac{7}{8}$
(D) $1 \frac{1}{4}$
(E) $\frac{3}{4}$
Suppose that the distance from Ginger's home to her school is $d \mathrm{~km}$.

Since there are 60 minutes in an hour, then $3 \frac{3}{4}$ minutes (or $\frac{15}{4}$ minutes) is $\frac{15}{4} \times \frac{1}{60}=\frac{1}{16}$ of an hour.

Since Ginger walks at $4 \mathrm{~km} / \mathrm{h}$, then it takes her $\frac{d}{4}$ hours to walk to school.

Since Ginger runs at $6 \mathrm{~km} / \mathrm{h}$, then it takes her $\frac{d}{6}$ hours to run to school.

Since she saves $\frac{1}{16}$ of an hour by running, then the difference between these times is $\frac{1}{16}$ of an hour, so

$$
\begin{aligned}
\frac{d}{4}-\frac{d}{6} & =\frac{1}{16} \\
\frac{3 d}{12}-\frac{2 d}{12} & =\frac{1}{16} \\
\frac{d}{12} & =\frac{1}{16} \\
d & =\frac{12}{16}=\frac{3}{4}
\end{aligned}
$$

Therefore, the distance from Ginger's home to her school is $\frac{3}{4} \mathrm{~km}$.

ANswer: (E)