A) 75.6 km/h
B) 70.80 km/h
C) 65.54 km/h
D) 45.25 km/h
Correct Answer: A
Solution :
Let the length of the train be x metres and its speed be y km/h i.e., \[\left( \frac{5y}{18} \right)\text{m/s}\] |
Then, \[\frac{x}{\left( \frac{5y}{18} \right)}=3\]\[\Rightarrow \]\[18x=15y\]\[\Rightarrow \]\[6x=5y\] |
Also, \[\frac{(x+105)}{\left( \frac{5y}{18} \right)}=8\]\[\Rightarrow \]\[18\,\,(x+105)=40\,\,y\] |
\[\Rightarrow \]\[9\,\,(x+105)=20y\]\[\Rightarrow \]\[20y-9x=945\] |
\[\Rightarrow \]\[24x-9x=945\] |
\[\Rightarrow \] \[15x=945\]\[\Rightarrow \]\[x=63\] |
\[\therefore \] \[5y=(6\times 63)\]\[\Rightarrow \]\[y=\frac{6\times 63}{5}=\frac{378}{5}=75.6\] |
Hence, the length of the train is 63 m and its speed is 75.6 km/h. |
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