A) 9 h
B) 7.5 h
C) 6.5 h
D) 8 h
Correct Answer: B
Solution :
If \[{{v}_{w}}\] be the velocity of water and \[{{v}_{b}}\] be the velocity of motorboat in still water. The distance covered by motorboat in moving downstream in 6 h is \[x=({{v}_{b}}+{{v}_{w}})\times 6\] ... (i) Same distance covered by motorboat in moving upstream in 10 h is \[x=({{v}_{b}}-{{v}_{w}})\times 10\] …. (ii) From Eqs. (i) and (ii), we have \[({{v}_{b}}+{{v}_{w}})\times 6=({{v}_{b}}-{{v}_{w}})\times 10\] \[{{v}_{w}}=\frac{{{v}_{b}}}{4}\] \[\therefore \] \[x=({{v}_{b}}+{{v}_{w}})\times 6=7.5\,{{v}_{b}}\] Time taken by the motorboat to cover the same distance in still water is \[t=\frac{x}{{{v}_{b}}}=\frac{7.5{{v}_{b}}}{{{v}_{b}}}=7.5\,h\]
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