At its usual speed, a 150 metre long train crosses a platform of lengt

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?1). 2502). 2253). 2404). 260
Answer» 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

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?1). 2502). 2253). 2404). 260

Answer»

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