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At 60% of its usual speed, a train of length L metres crosses a platform 240 metre long in 15 seconds. At its usual speed, the train crosses a pole in 6 seconds. What is the value of L (in metre)?
- A270
- B225
- C220
- D480
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Solution
Let speed of the train = $10x$ m/s
Length of train = $l$ m
Time taken to cross the pole = 6 sec
Using, $speed = \frac{distance}{time}$
=> $10x = \frac{l}{6}$
=> $x = \frac{l}{60}$
Now, 60% of the speed = $\frac{60}{100} \times 10x = 6x$ m/s
Length of platform = 240 m
Acc. to ques, => $6x = \frac{240 + l}{15}$
=> $6 \times \frac{l}{60} = \frac{240 + l}{15}$
=> $\frac{l}{10} = \frac{240 + l}{15}$
=> $15l = 2400 + 10l$
=> $15l - 10l = 5l = 2400$
=> $l = \frac{2400}{5} = 480$ m
Let speed of the train = $10x$ m/s
Length of train = $l$ m
Time taken to cross the pole = 6 sec
Using, $speed = \frac{distance}{time}$
=> $10x = \frac{l}{6}$
=> $x = \frac{l}{60}$
Now, 60% of the speed = $\frac{60}{100} \times 10x = 6x$ m/s
Length of platform = 240 m
Acc. to ques, => $6x = \frac{240 + l}{15}$
=> $6 \times \frac{l}{60} = \frac{240 + l}{15}$
=> $\frac{l}{10} = \frac{240 + l}{15}$
=> $15l = 2400 + 10l$
=> $15l - 10l = 5l = 2400$
=> $l = \frac{2400}{5} = 480$ m