-1
At its usual speed, a 150 metre long train crosses a platform of length L metres in 24 seconds. AT 75% of its usual speed, the train crosses a vertical pole in 12 seconds. What is the value of L?
- A250
- B225
- C240
- D260
Answer not set
Discussion 1
Log in or create a free account to reply and vote.
0
CCommunity
Solution
Let speed of the train = $20x$ m/s
Now, 75% of the speed = $\frac{75}{100} \times 20x = 15x$ m/s
Length of train = 150 m
Time taken to cross the pole = 12 sec
Using, $speed = \frac{distance}{time}$
=> $15x = \frac{150}{12}$
=> $x = \frac{10}{12} = \frac{5}{6}$
Length of platform = $l$ m
Acc. to ques, => $20x = \frac{150 + l}{24}$
=> $20 \times \frac{5}{6} = \frac{150 + l}{24}$
=> $\frac{50}{3} = \frac{150 + l}{24}$
=> $150 + l = 400$
=> $l = 400 - 150 = 250$ m
Let speed of the train = $20x$ m/s
Now, 75% of the speed = $\frac{75}{100} \times 20x = 15x$ m/s
Length of train = 150 m
Time taken to cross the pole = 12 sec
Using, $speed = \frac{distance}{time}$
=> $15x = \frac{150}{12}$
=> $x = \frac{10}{12} = \frac{5}{6}$
Length of platform = $l$ m
Acc. to ques, => $20x = \frac{150 + l}{24}$
=> $20 \times \frac{5}{6} = \frac{150 + l}{24}$
=> $\frac{50}{3} = \frac{150 + l}{24}$
=> $150 + l = 400$
=> $l = 400 - 150 = 250$ m