At 60% of its usual speed, a train of length L metres crosses a platfo

1.	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)?1). 2702). 2253). 2204). 480
Answer» 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

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)?1). 2702). 2253). 2204). 480

Answer»

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