| A river is flowing with a steady speed of 4 km/h. One rows his boat downstream in the river and then returns by rowing upstream in the same river. When he returns to the starting point, the total distance covered by him is 42 km. If the return journey takes 2 h more than his outward journey, then the speed of his rowing in still water must be |
A) 12 km/h
B) 10 km/h
C) 9 km/h
D) 8 km/h
Correct Answer: B
Solution :
| Let the speed of rowing in still water be \[u\,\,km/h.\] |
| Distance in downstream motion \[=21\,\,km\] |
| and speed downstream \[=(u+4)\,\,km/h\] |
| \[\therefore \] Time taken \[=\frac{21}{u+4}\] |
| Distance upstream motion \[=21\,\,km\] |
| and speed upstream \[=(u+4)\,\,km\] |
| \[\therefore \] Time taken \[=\frac{21}{u-4}\] |
| According to the question, |
| \[\frac{21}{u+4}+2=\frac{21}{u-4}\] |
| \[\Rightarrow \] \[\frac{21+2u+8}{u+4}=\frac{21}{u-4}\] |
| \[\Rightarrow \] \[\frac{2u+29}{u+4}=\frac{21}{u-4}\] |
| \[\Rightarrow \] \[(u-4)(2u+29)=21\,\,(u+4)\] |
| \[\Rightarrow \]\[2{{u}^{2}}-8u+29u-116=21u+84\] |
| \[\Rightarrow \] \[2{{u}^{2}}+21u-116=21u+84\] |
| \[\Rightarrow \] \[2{{u}^{2}}=84+116\] |
| \[\Rightarrow \] \[{{u}^{2}}=\frac{200}{2}\]\[\Rightarrow \]\[{{u}^{2}}=100\] |
| \[\Rightarrow \] \[u=10km/h\] |
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