| A train running at \[8/11\,\,km/h\] of its own speed reached a place in \[5\frac{1}{2}h.\]How much time could be saved, if the train would have run at it own speed? |
A) \[2\frac{1}{2}h\]
B) \[2\,\,h\]
C) \[1\frac{1}{2}h\]
D) \[1\,\,h\]
Correct Answer: C
Solution :
(c)| New speed \[=\frac{8}{11}\]of usual speed. |
| \[\therefore \]New time \[=\frac{11}{8}\]of. usual time. |
| So, \[\frac{11}{8}\]of usual time \[=\frac{11}{2}\text{h}\] |
| \[\Rightarrow \]Usual time \[=\frac{11\times 8}{2\times 11}=4\,\,\text{h}\] |
| Hence, time saved \[=5\frac{1}{2}-4=1\frac{1}{2}\text{h}\]. |
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