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 \[t=\frac{x}{{{v}_{b}}}=\frac{7.5{{v}_{b}}}{{{v}_{b}}}=7.5h\]
You need to login to perform this action.
You will be redirected in
3 sec
