The problem requires calculating the average speed for a journey covering the same distance twice, but at different speeds.
Average Speed = \(\frac{2}{\frac{1}{13 \, \text{km/h}} + \frac{1}{10 \, \text{km/h}}}\)
Average Speed = \(\frac{2}{\frac{10}{130} + \frac{13}{130}}\)
Average Speed = \(\frac{2}{\frac{10 + 13}{130}}\)
Average Speed = \(\frac{2}{\frac{23}{130}}\)
Average Speed = \(2 \times \frac{130}{23}\)
Average Speed = \(\frac{260}{23}\) km/h
\(260 \div 23 = 11\) with a remainder of \(7\).
Therefore, the average speed is \(11\frac{7}{23}\) km/h.
The average speed for the whole journey is \(\frac{260}{23}\) km/h, which is equivalent to \(11\frac{7}{23}\) km/h.
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