What is the speed of the train the train crosses a signal pole in 18 seconds the train crosses a platform of equal length in 36 seconds length of the train is 330 m?

In what time will a train 100 meters long cross an electric pole, if its speed be 144 kmph?

Level 1 Problems on Trains Formula Based Questions

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A

Solution

\(Speed=\left ( 144*\frac{5}{18} \right )m/sec=40m/sec\)

time taken=(100/40)sec=2.5 sec

A train 360 m long is running at a speed of 45 kmph. In what time will it pass a bridge 140 m long?

Level 1 Problems on Trains Formula Based Questions

Your response is correct

Your response is incorrect

A

Solution

\(speed=\left ( 45*\frac{5}{18} \right )m/sec=\frac{25}{2}m/sec\)

Total distance covered = (360+140)m = 500 m

\(Required\, time=\left ( 500*\frac{2}{25} \right )sec=40\, sec\)

A train running at the speed of 60 kmph crosses a pole in 9 seconds. What is the length of the train?

Level 1 Problems on Trains Formula Based Questions

Your response is correct

Your response is incorrect

E

Solution

\(Speed=\left ( 60*\frac{5}{18} \right )m/sec=\left ( \frac{50}{3} \right )m/sec\)

Length of the train = (Speed * Time) \(=\left ( \frac{50}{3}*9 \right )m=150\, m\)

The length of the bridge, which a train 130 meters long and travelling at 45 kmph can cross in 30 seconds, is:

Level 1 Problems on Trains Relative Motion of Train

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C

Solution

\(Speed=\left ( 45*\frac{5}{18} \right )m/sec=\left ( \frac{25}{2} \right )m/sec;\) Time = 30 sec

Let the length of bridge be x metres.

Then, \(\frac{130+x}{30}=\frac{25}{2}\)

2(130+x) = 750

x= 245 m

A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

Level 1 Problems on Trains

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Your response is incorrect

B

Solution

\(speed=\left ( \frac{300}{18} \right )m/sec=\frac{50}{3}m/sec\)

Let the length of the platform be x metres.

Then, \(\frac{x+300}{39}=\frac{50}{3}\)

3(x+300)=1950

x = 350

A train 800 metres long is running at a speed of 78 kmph. If it crosses a tunnel in 1 minute, then the length of the tunnel (in metres) is:

Level 1 Problems on Trains Relative Speed of Train

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C

Solution

\(Speed=\left ( 78*\frac{5}{18} \right )m/sec=\left ( \frac{65}{3} \right )m/sec\)

Time = 1 minute = 60 sec

Let the length of the tunnel be x meters.

Then, \(\frac{800+x}{60}=\frac{65}{3}\)

3(800+x) = 3900

x = 500

A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

Level 1 Problems on Trains Relative Motion of Train

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C

Solution

Speed of train relative to jogger = (45-9) kmph = 36 kmph

\(=\left ( 36*\frac{5}{18} \right )m/sec=10\, m/sec\)

Distance to be covered = (240+120) m = 360 m

Time taken = (\(\frac{360}{10}\)) = 36 sec

Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

Level 1 Problems on Trains Relative Motion of Train

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C

Solution

Let the speed of the slower train be x m/sec

Then, speed of the fastre train = 2x m/sec

Relative speed = (x+2x) m/sec = 3x m/sec

\(\frac{(100+100)}{8}=3x\)

24x = 200

x = \(\frac{25}{3}\)

So, speed of the fastre train = \(\frac{50}{3}m/sec=\left ( \frac{50}{3}*\frac{18}{5} \right )kmph=60kmph\)

Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Level 2 Problems on Trains Relative Motion of Train

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B

Solution

Let us name the train as A and B. Then

(A's speed):(B's speed) = \(\sqrt{b}:\sqrt{a}=\sqrt{16}:\sqrt{9}=4:3\)

A train crosses a pole in 10 seconds. What is the length of the train?

I. The train crosses another train running in opposite direction with a speed of 80 km/hr in 22 seconds.

II. The speed of the train is 108 kmph.

Level 2 Problems on Trains Relative Speed of Train

Your response is correct

Your response is incorrect

B

Solution

Time taken to cross a pole = \(\frac{Length\, of\, train}{Speed\, of\, train}=10\Rightarrow \frac{Length\, of\, train}{\left (108*\frac{5}{18} \right )}\)

Length of the train = 300 m

Clearly, II is sufficient to get the answer.

Also, I is not sufficient to get the answer.

What is the speed of the train?

I. The train crosses a signal pole in 18 seconds.

II. The train crosses a platform of equal length in 36 seconds

III. Length of the train is 330 metres.

Level 2 Problems on Trains Relative Speed of Train

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Your response is incorrect

D

Solution

Time taken to cross a signal pole = \(=\frac{length \, of\, the\, train}{speed \, of\, the\, train}\)

Time taken to cross a platform = \(=\frac{(length \, of\, the\, train+length \, of\, the\, platform)}{speed\, of\, the\, train}\)

Length of the train = 330 m

I and III give, \(18=\frac{330}{x}\Rightarrow x=\frac{330}{18}m/s=\frac{55}{3}m/s\)

II and III give, \(36=\frac{2*330}{x}\Rightarrow x=\frac{660}{36}m/s=\frac{55}{3}m/s\)

At what time will the train reach city X from city Y?

I. The train crosses another train of equal length of 200 metres and running in opposite directions in 15 seconds

II. The train leaves city Y at 7.15 a.m. for city X situated at a distance of 558 km.

III. The 200 metres long train crosses a signal pole in 10 seconds.

Level 2 Problems on Trains Relative Motion of Train

Your response is correct

Your response is incorrect

D

Solution

III. Gives, speed = \(\frac{200}{10}m/s=20m/s=\left ( 20*\frac{18}{5} \right )kmph=72\, kmph\)

II gives, time taken = \(\left ( \frac{558}{72} \right )hrs=\frac{31}{4}hrs=7\frac{3}{4}hrs=7\, hrs\, 45\, min\)

So, the train will reach city X at 3 p.m.

Hence, I is redundant.

A man sitting in a train is counting the pillars of electricity. The distance between two pillars is 60 m, and the speed of the train is 42 km/h. In 5 hours, how many pillars will he count?

Level 2 Problems on Trains Formula Based Questions

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A

Solution

Distance covered by the train in 5 hours = 42*5 = 210 km = 210000 m

Number of pillars counted by the man = \(\frac{210000}{60}+1\) = 3500+1 = 3501

A train takes 5 minutes to cross a telegraphic post. Then the time taken by another train whose length is just double of the first train and moving with the same speed to cross a platform of its own length is

Level 2 Problems on Trains Relative Speed of Train

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C

Solution

Let the length of the train be x metres.

Time taken to cover x metres = 5 min = 5*60 = 300 sec

Speed of the train = \(\frac{x}{300}\) m/sec

Length of the second train = 2x metres

Length of the platform = 2x metres

Required time = \(\frac{2x+2x}{\frac{x}{300}}\) = \(\frac{4x*300}{x}\) = 1200 sec = \(\frac{1200}{60}\) min = 20 min

A train with 90 km/hr crosses a bridge in 36 seconds. Another train 100 metres shorter crosses the same bridge at 45 km/h. What is the time taken by the second train to cross the bridge?

Level 2 Problems on Trains Relative Motion of Train

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D

Solution

Let the lengths of the train and the bride be x and y m respectively

Speed of the first train = 90 km/h = 90*\(\frac{5}{18}\)  = 25 m/s

Speed of the second train = 45 km/h = 45*\(\frac{5}{18}\)  = \(\frac{25}{2}\) m/s

Then, \(\frac{x+y}{36}\) =25

x+y=900

Required time = \(\frac{x-100+y}{25/2}\) = \(\frac{x+y-100}{25/2}\)  

800*\(\frac{2}{25}\) = 64 sec 

The Ghaziabad-Hapur-Meerut EMU and the Meerut-Hapur-Ghaziabad EMU start at the same time from Ghaziabad and Meerut and procees towards each other at 16 km/hr and 21 km/hr respectively. When they meet, it is found that one train has traveled 60 km more than the other. The distance between two stations is

Level 2 Problems on Trains Relative Motion of Train

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Your response is incorrect

B

Solution

At the time of meeting,let the distance travelled by the first train be x km

Then, distance traveled by the second train is (x+60) km

\(\frac{x}{16}\) = \(\frac{x+60}{21}\)

21x=16x+960

5x=960

x=192

Hence, distance between two stations = 192+192+60 = 444 km

Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have traveled 175 km more than the other. Find the distance between A and B.

Level 3 Problems on Trains Relative Motion of Train

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Your response is incorrect

A

Solution

Let the train meet after t hours

Speed of train A = 75 km/h

Speed of train B = 50 km/h

Distance covered by train A = 75*t = 75t

Distance covered by train B = 50*t = 50t

Distance = Speed*Time

According to question,

75t-50t = 175

25t=175

t=\(\frac{175}{25}\) = 7 hours

Distance between A and B = 75t+50t=125t

125*7=875 km

Two trains start simultaneously from two stations 270 km apart, each to the opposite station; they reach their destinations in 6\(\frac{1}{4}\) hours and 4 hours after they meet. The rate at which the slower train travels is

Level 3 Problems on Trains Relative Motion of Train

Your response is correct

Your response is incorrect

B

Solution

Ratio of speeds = \(\sqrt{4}\) : \(\sqrt{6\frac{1}{4}}\) : \(\sqrt{4}\) : \(\sqrt{\frac{25}{4}}\) = 2 : \(\frac{5}{2}\) = 4:5

Let the speed of the two trains be 4 x and 5 x km/h respectively.

Then, time taken by trains to meet each other = \(\frac{270}{4x+5x}\) = \(\frac{270}{9x}\) = \(\frac{30}{x}\)

Time taken by slower train to travel 270 km = \(\frac{270}{4x}\) hr

\(\frac{270}{4x}\)=\(\frac{30}{x}\)+6\(\frac{1}{4}\)

\(\frac{270}{4x}\)-\(\frac{30}{x}\)=\(\frac{25}{4}\)

\(\frac{150}{4x}\) = \(\frac{25}{4}\)

100x=600

x=6

Hence, speed of slower train = 4x=24 km/h

Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was 17 km, and another 15 minutes later, one sea trawler was 10.5 km farther from the origin than the other. Find the speed of each sea trawler.

Level 3 Problems on Trains Relative Motion of Train

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Your response is incorrect

A

Solution

Suppose the two trawlers start from a point O and move inn the directions OA and OB respectively.

Let the speeds of the two sea trawlers be x km/hr and y km/hr respectively.

Then, [x*\(\frac{1}{2}]^{2}\)+ [y*\(\frac{1}{2}]^{2}\) = \(17^{2}\)

 \(\frac{x^{2}}{4}\)+ \(\frac{y^{2}}{4}\)=289

\(x^{2}+y^{2}\)=1156           ..i

And, x* \(\frac{3}{4}\)-y*\(\frac{3}{4}\)=10.5

x-y=10.5*\(\frac{4}{3}\)=14       ...ii

Now, [\(x+y]^{2}\)+[\(x-y]^{2}\)=2[\(x^{2}+y^{2}\)]

[\(x+y]^{2}\)=2*1156-\(14^{2}\) = 2312-196 = 2166

x+y=\(\sqrt{2116}\) = 46     ..iii

Adding ii and iii, we get: 2x=60 or x=30

Putting x=30 in ii we get, y= 16

Hence, the speeds of the two sea-trawlers are 30 km/hr and 16 km/hr

Two trains start simultaneously from two stations 270 km apart, each to the opposite station; they reach their destinations in 6\(\frac{1}{4}\) hours and 4 hours after they meet. The rate at which the slower train travels is

Level 3 Problems on Trains Relative Motion of Train

Your response is correct

Your response is incorrect

B

Solution

Ratio of speeds = \(\sqrt{4}\) : \(\sqrt{6\frac{1}{4}}\) : \(\sqrt{4}\) : \(\sqrt{\frac{25}{4}}\) = 2 : \(\frac{5}{2}\) = 4:5

Let the speed of the two trains be 4 x and 5 x km/h respectively.

Then, time taken by trains to meet each other = \(\frac{270}{4x+5x}\) = \(\frac{270}{9x}\) = \(\frac{30}{x}\)

Time taken by slower train to travel 270 km = \(\frac{270}{4x}\) hr

\(\frac{270}{4x}\)=\(\frac{30}{x}\)+6\(\frac{1}{4}\)

\(\frac{270}{4x}\)-\(\frac{30}{x}\)=\(\frac{25}{4}\)

\(\frac{150}{4x}\) = \(\frac{25}{4}\)

100x=600

x=6

Hence, speed of slower train = 4x=24 km/h