A) \[0.4\]
B) 0.2
C) 1
D) \[0.6\]
Correct Answer: D
Solution :
The distance covered upstream \[=AC=d\] \[AB=100\] \[BC=100+d\] Rate upstream = \[(x-y)\] m/minute Rate downstream =\[(x+y)\] m/minute \[\therefore \] \[\frac{d}{x-y}=5\] \[\Rightarrow \] \[d=5\,\,(x-y)\] ?..(i) Again, \[\frac{100+d}{x+y}=5\] \[\Rightarrow \] \[\frac{100+5(x-y)}{x+y}=5\] \[\Rightarrow \] \[100+5x-5y=5x+5y\] \[\Rightarrow \] \[10y=100\] \[\Rightarrow \] \[y=10\,m/\text{minute}\] \[=\frac{10}{1000}\times 60\,\,kmph\] \[=0.6\,\,kmph\]You need to login to perform this action.
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