Let's solve this problem by using vectors. We can represent the girl's displacement from point Q to point X as a vector \(\vec{QX}\) and her displacement from point X to point Y as a vector \(\vec{XY}\). The magnitude of \(\vec{QX}\) is 45 meters and the magnitude of \(\vec{XY}\) is 24 meters. The direction of \(\vec{QX}\) is 050° and the direction of \(\vec{XY}\) is 140°. We can represent these vectors in rectangular coordinates as follows:

\[\vec{QX} = 45\begin{pmatrix} \cos 050° \\ \sin 050° \end{pmatrix} = \begin{pmatrix} 45\cos 050° \\ 45\sin 050° \end{pmatrix}\]
\[\vec{XY} = 24\begin{pmatrix} \cos 140° \\ \sin 140° \end{pmatrix} = \begin{pmatrix} 24\cos 140° \\ 24\sin 140° \end{pmatrix}\]

The girl's total displacement from point Q to point Y is the sum of these two vectors: \(\vec{QY} = \vec{QX} + \vec{XY}\). So we have:

\[\vec{QY} = \begin{pmatrix} 45\cos 050° \\ 45\sin 050° \end{pmatrix} + \begin{pmatrix} 24\cos 140° \\ 24\sin 140° \end{pmatrix} = \begin{pmatrix} 45\cos 050° + 24\cos 140° \\ 45\sin 050° + 24\sin 140° \end{pmatrix}\]

The distance from point Q to point Y is the magnitude of the vector \(\vec{QY}\), which is given by:

\[|\vec{QY}| = \sqrt{(45\cos 050° + 24\cos 140°)^2 + (45\sin 050° + 24\sin 140°)^2}\]

Using a calculator, we find that this expression evaluates to approximately 51 meters. So the girl is 51 meters away from point Q.