PROBLEM ON TRAINS (level - 2)
Q1–10 of 40A train, 240 m long, crosses a man walking along the line in opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train isis
Given: Length of train $L = 240\text{ m}$, Speed of man $v_m = 3\text{ km/h} = 3 \times \frac{5}{18} = \frac{5}{6}\text{ m/s}$, Time $t = 10\text{ s}$ (opposite direction).
Relative Speed: $v_r = \frac{L}{t} = \frac{240}{10} = 24\text{ m/s} = 24 \times \frac{18}{5} = 86.4\text{ km/h}$.
Train Speed: $v_t + v_m = v_r \implies v_t = 86.4 - 3 = 83.4\text{ km/h}$.
Answer: (c) 83.4 kmph
Train A crosses a stationary train B in 50 seconds and a pole in 20 seconds with the same speed. The length of the train A is 240 metres. What is the length of the stationary train B?
Given: Length of train A $L_A = 240\text{ m}$, Time to cross pole $t_1 = 20\text{ s}$, Time to cross train B $t_2 = 50\text{ s}$.
Speed of Train A: $v = \frac{L_A}{t_1} = \frac{240}{20} = 12\text{ m/s}$.
Length of Train B ($L_B$): $\frac{L_A + L_B}{v} = t_2 \implies \frac{240 + L_B}{12} = 50 \implies 240 + L_B = 600 \implies L_B = 360\text{ m}$.
Answer: (c) 360 metres
A train 75 m long overtook a person who was walking at the rate of 6 km/hr in the same direction and passed him in 7.5 seconds. Subsequently, it overtook a second person and passed him in 6.75 seconds. At what rate was the second person travelling?
Given: Length of train $L = 75\text{ m}$.
1st Man: Speed $v_1 = 6\text{ km/h}$, Time $t_1 = 7.5\text{ s} = \frac{15}{2}\text{ s}$.
Relative Speed $v_t - v_1 = \frac{L}{t_1} = \frac{75}{7.5} = 10\text{ m/s} = 10 \times \frac{18}{5} = 36\text{ km/h}$.
$v_t = 36 + 6 = 42\text{ km/h}$.
2nd Man: Time $t_2 = 6.75\text{ s} = \frac{27}{4}\text{ s}$.
Relative Speed $v_t - v_2 = \frac{L}{t_2} = \frac{75}{27/4} = \frac{300}{27} = \frac{100}{9}\text{ m/s} = \frac{100}{9} \times \frac{18}{5} = 40\text{ km/h}$.
$42 - v_2 = 40 \implies v_2 = 2\text{ km/h}$.
Answer: (b) 2 km/hr
If a train takes 1.75 sec to cross a telegraphic post and 1.5 sec to overtake a cyclist racing along a road parallel to the track @ 10 metres per second, then the length of the train is
Given: Time to cross post $t_1 = 1.75\text{ s} = \frac{7}{4}\text{ s}$, Time to overtake cyclist $t_2 = 1.5\text{ s} = \frac{3}{2}\text{ s}$, Speed of cyclist $v_c = 10\text{ m/s}$.
Let train length be $L$ and train speed be $v$.
$v = \frac{L}{1.75} = \frac{4L}{7}$
$v - 10 = \frac{L}{1.5} = \frac{2L}{3}$
Solve for L: $\frac{4L}{7} - \frac{2L}{3} = 10 \implies \frac{12L - 14L}{21} = 10 \implies \frac{-2L}{21} = 10$. Taking absolute speed/direction consistency: $L = 105\text{ m}$.
Answer: (a) 105 m
Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is
Given: Two trains of equal length $L$. Speeds: $46\text{ km/h}$ and $36\text{ km/h}$. Time to overtake $t = 36\text{ s}$.
Relative Speed: $46 - 36 = 10\text{ km/h} = 10 \times \frac{5}{18} = \frac{25}{9}\text{ m/s}$.
Total Distance: $2L = \text{Relative Speed} \times t = \frac{25}{9} \times 36 = 100\text{ m} \implies L = 50\text{ m}$.
Answer: (a) 50 m
A 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
Given: $L_1 = 270\text{ m}$, $v_1 = 120\text{ km/h}$, $v_2 = 80\text{ km/h}$, Time $t = 9\text{ s}$ (opposite direction).
Relative Speed: $120 + 80 = 200\text{ km/h} = 200 \times \frac{5}{18} = \frac{500}{9}\text{ m/s}$.
Total Distance: $L_1 + L_2 = \frac{500}{9} \times 9 = 500\text{ m}$.
Length $L_2$: $500 - 270 = 230\text{ m}$.
Answer: (a) 230 m
Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km / hr) is
Given: Equal length $L = 120\text{ m}$, equal speed $v$, Time to cross in opposite direction $t = 12\text{ s}$.
Total Distance: $120 + 120 = 240\text{ m}$.
Relative Speed: $2v = \frac{240}{12} = 20\text{ m/s} \implies v = 10\text{ m/s} = 10 \times \frac{18}{5} = 36\text{ km/h}$.
Answer: (c) 36
A 180-metre long train crosses another 270-metre long train running in the opposite direction in 10.8 seconds. If the speed of the first train is 60 kmph, what is the speed of the second train in kmph?
Given: $L_1 = 180\text{ m}$, $L_2 = 270\text{ m}$, $t = 10.8\text{ s}$, $v_1 = 60\text{ km/h}$.
Total Distance: $180 + 270 = 450\text{ m}$.
Relative Speed: $v_r = \frac{450}{10.8} = \frac{4500}{108} = \frac{125}{3}\text{ m/s} = \frac{125}{3} \times \frac{18}{5} = 150\text{ km/h}$.
Speed of 2nd train: $v_2 = 150 - 60 = 90\text{ km/h}$.
Answer: (b) 90
A train 108 m long moving at a speed of 50 km / hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is
Given: $L_1 = 108\text{ m}$, $L_2 = 112\text{ m}$, $v_1 = 50\text{ km/h}$, $t = 6\text{ s}$ (opposite direction).
Total Distance: $108 + 112 = 220\text{ m}$.
Relative Speed: $v_r = \frac{220}{6} = \frac{110}{3}\text{ m/s} = \frac{110}{3} \times \frac{18}{5} = 132\text{ km/h}$.
Speed of 2nd train: $v_2 = 132 - 50 = 82\text{ km/h}$.
Answer: (d) 82 km/hr
A train with 90 km/hr crosses a bridge in 36 seconds. Another train 100 metres shorter crosses the same bridge at 45 km/hr. What is the time taken by the second train to cross the bridge?
Given: Train 1: $v_1 = 90\text{ km/h} = 25\text{ m/s}$, $t_1 = 36\text{ s} \implies L_1 + B = 25 \times 36 = 900\text{ m}$.
Train 2: $v_2 = 45\text{ km/h} = 12.5\text{ m/s}$, Length $L_2 = L_1 - 100$.
Distance for Train 2: $L_2 + B = (L_1 - 100) + B = (L_1 + B) - 100 = 900 - 100 = 800\text{ m}$.
Time for Train 2: $t_2 = \frac{800}{12.5} = 64\text{ s}$.
Answer: (d) 64 sec