Answer:
Let the distance covered by the cyclists be ‘d’. Let \[{{V}_{1}}\]be the lower speed and \[{{V}_{2}}\]be the higher speed.
\[\therefore {{v}_{1}}=\frac{d}{50}\] ………(1)
\[{{v}_{2}}=\frac{d}{40}\] ………(2)
Case 1:- Starts with lower speed \[({{v}_{1}}).\]
Distance covered in first ten minutes with lower
speed \[{{v}_{1}}=\frac{d}{50}\times 10=\frac{d}{5}\]
Distance covered in next ten minutes with higher speed\[{{v}_{2}}=\frac{d}{40}\times 10=\frac{d}{4}\]
Hence, in fourty minutes he covers
\[=\frac{d}{5}+\frac{d}{4}+\frac{d}{5}+\frac{d}{4}=\frac{9d}{10}\]
Distance remaining \[=d-\frac{9d}{10}=\frac{d}{10}\]
He covers this distance with lower speed\[{{V}_{1}}\]in
\[\frac{d/10}{d/50}=5\,\min \]
\[\therefore \]Total time taken\[=40+5=45\,\,\min .\]
For case 2, when he starts with higher speed we get 44 min.
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