Problems on Trains - Aptitude Questions, Shortcuts, Formulas

We cover all the topics and concepts of Aptitude Problems on trains like two trains are moving in opposite directions, calculate train speed, calculate distance, etc.

Rules on Problem On Trains

Rule 1: When two trains are moving in opposite directions, then relative speed will be the addition of their individual speeds.

Rule 2: When two trains are moving in same direction, then relative speed will be the subtraction of their individual speeds.

Rule 3: On passing a platform by a certain train the net distance traveled is the sum of length of train and the length of platform both.

Rule 4: When a train passes through a pole or person standing, net distance traveled to pass is the length of the train

Problems on Trains Formulas

x $km/hr = x \times \frac{5}{18} m/s$
y $m/s = y\times \frac{18}{5} Km/hr$
Speed = distance/time
Velocity = displacement/time
Time taken by a train x meters long to pass a pole or standing man or a post = Time taken by the train to travel x meters.
Time taken by a train x meters long to pass an object of length y meters = Time taken by the train to travel (x + y) metres.
Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v 1 > v 2, then their relative speed = (v 1 v 2) m/s
Suppose two trains or two objects are moving in opposite directions at v 1 m/s and v 2 m/s , then their relative speed = (v 1+ v 2) m/s
Assume two trains of length x metres and y metres are moving in opposite directions at v 1 m/s and v 2 m/s, Then the time taken by the trains to cross each other = (x+y) / (v 1+v 2) seconds
Assume two trains of length x metres and y metres are moving in the same direction at at v 1 m/s and v 2 m/s where v 1 > v 2, Then The time taken by the faster train to cross the slower train = (x+y) / (v 1-v 2) seconds.
Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then, A's speed: B's speed = $\sqrt{q} : \sqrt{p}$

Problems on Trains Aptitude Questions from Previous Year Exams

IBPS questions related to problem on train.

Question 1 of 33

1. Question
1 points
Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?
- 23 m
- 250/9 m
- 23.22 m
- 29 m
Correct

Relative speed = (40 - 20) km/hr

$=\left ( 20\times \frac{5}{18} \right )m/sec=\left ( \frac{50}{9} \right )m/sec$

Length of faster train = $\left ( 5\times \frac{50}{9} \right )m=\left ( \frac{250}{9} \right )m$

Incorrect

Relative speed = (40 - 20) km/hr

$=\left ( 20\times \frac{5}{18} \right )m/sec=\left ( \frac{50}{9} \right )m/sec$

Length of faster train = $\left ( 5\times \frac{50}{9} \right )m=\left ( \frac{250}{9} \right )m$
Question 2 of 33

2. Question
1 points
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:
- 10km/hr
- 36km/hr
- 18km/hr
- 72km/hr
Correct

Let the speed of each train be x m/sec.

Then, relative speed of the two trains = 2x m/sec.

So 2x = $\frac{\left ( 120+120 \right )}{12}$

2x = 20

x = 10.

Speed of each train = 10 m/sec = $\left ( 10\times \frac{18}{5} \right )$ km/hr = 36 km/hr

Incorrect

Let the speed of each train be x m/sec.

Then, relative speed of the two trains = 2x m/sec.

So 2x = $\frac{\left ( 120+120 \right )}{12}$

2x = 20

x = 10.

Speed of each train = 10 m/sec = $\left ( 10\times \frac{18}{5} \right )$ km/hr = 36 km/hr
Question 3 of 33

3. Question
1 points
A train takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is
- 70 m
- 80 m
- 90 m
- 100 m
Correct

Let the length of the train be x meters. = $\frac{x+162}{15}$

= $\frac{x+120}{15}$

= 15( x+162)

= 18(x+120)

x= 90.

Incorrect

Let the length of the train be x meters. = $\frac{x+162}{15}$

= $\frac{x+120}{15}$

= 15( x+162)

= 18(x+120)

x= 90.
Question 4 of 33

4. Question
1 points
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?
- 320 m
- 350 m
- 650 m
- Data inadequate
Correct

Speed = $\left(\frac{300}{18}\right)$ m/sec = 50 m/sec.

Let the length of the platform be x meters. Then, $\left[\frac{(x+300)}{39}\right]=\frac{50}{3}$

=> 3(x + 300) = 1950

=> x = 350 m.

Incorrect

Speed = $\left(\frac{300}{18}\right)$ m/sec = 50 m/sec.

Let the length of the platform be x meters. Then, $\left[\frac{(x+300)}{39}\right]=\frac{50}{3}$

=> 3(x + 300) = 1950

=> x = 350 m.
Question 5 of 33

5. Question
1 points
Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?
- 10
- 12
- 15
- 20
Correct

Speed of the first train = $\frac{120}{10}$ m/sec = 12 m/sec.

Speed of the second train = $\frac{120}{15}$ m/sec = 8 m/sec.

Relative speed = (12 + 8) = 20 m/sec.

Therefore Required time = $\left[\frac{(120+120)}{20}\right]$ sec = 12 sec.

Incorrect

Speed of the first train = $\frac{120}{10}$ m/sec = 12 m/sec.

Speed of the second train = $\frac{120}{15}$ m/sec = 8 m/sec.

Relative speed = (12 + 8) = 20 m/sec.

Therefore Required time = $$\left[\frac{(120+120)}{20}\right]$ sec = 12 sec.
Question 6 of 33

6. Question
1 points
A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
- 45 m
- 54 m
- 50 m
- 72 m
Correct

Incorrect

$2\, kmph=\left ( 2\times \frac{5}{18} \right )m/sec=\frac{5}{9}m/sec$

$4\, kmph=\left ( 4\times\frac{5}{18} \right )m/sec=\frac{10}{9}m/sec$

Let the length of the train be x metres and its speed by y m/sec

$Then,\, \left (\frac{x}{y-\frac{5}{9}} \right )=9 \, \, and\, \,\left (\frac{x}{y-\frac{10}{9}} \right )=10$

9y - 5 = x and 10(9y - 10) = 9x

9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Length of the train is 50 m.
Question 7 of 33

7. Question
1 points
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
- 45 km/hr
- 54 km/hr
- 50 km/hr
- 55 km/hr
Correct

Incorrect

Speed of the train relative to man = $\left ( \frac{125}{10} \right )m/sec$

$\left ( \frac{25}{2} \right )m/sec$

$\left ( \frac{25}{2}\times \frac{18}{5} \right )km/h=45 km/hr$

Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.

x - 5 = 45
x = 50 km/hr.
Question 8 of 33

8. Question
1 points
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
- 120 metres
- 324 metres
- 180 metres
- 150 metres
Correct

Incorrect

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

Length of the train = (Speed x Time)

$\left (\frac{50}{3} \times 9\right )m=150m$
Question 9 of 33

9. Question
1 points
A 270 metres 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?
- 230 m
- 260 m
- 240 m
- 270 m
Correct

Relative speed = (120 + 80) km/hr

$\left ( 200\times \frac{5}{18} \right )m/sec$

$\left (\frac{500}{9}\right )m/sec$

Let the length of the other train be x metres.

Then, $\frac{x+270}{9}=\frac{500}{9}$

x+ 270 = 500

x = 230.

Incorrect

Relative speed = (120 + 80) km/hr

$\left ( 200\times \frac{5}{18} \right )m/sec$

$\left (\frac{500}{9}\right )m/sec$

Let the length of the other train be x metres.

Then, $\frac{x+270}{9}=\frac{500}{9}$

x+ 270 = 500

x = 230.
Question 10 of 33

10. Question
1 points
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:
- 30 km/hr
- 60 km/hr
- 45 km/hr
- 75 km/hr
Correct

Incorrect

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

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

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

$\therefore \frac{100+100}{8}=3x$

24x = 200

$\Rightarrow x=\frac{25}{3}$

So, the speed of the faster train= $\frac{50}{3}m/sec$

$\left (\frac{50}{3}\times\frac{18}{5}\right )km/h= 60km/h$
Question 11 of 33

11. Question
1 points
Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?
- 10
- 15
- 12
- 20
Correct

$Speed\, of \,the\,first\,train =\frac{120}{10}m/sec=12\, m/sec$

$Speed\,of\,the\, second\,train =\frac{120}{15}m/sec=8\, m/sec$

Relative speed = (12 + 8) =20 m/sec.

Required time = $\left [ \frac{\left ( 120+120 \right )}{20} \right ]sec=12\,sec$

Incorrect

$Speed\, of \,the\,first\,train =\frac{120}{10}m/sec=12\, m/sec$

$Speed\,of\,the\, second\,train =\frac{120}{15}m/sec=8\, m/sec$

Relative speed = (12 + 8) =20 m/sec.

Required time = $\left [ \frac{\left ( 120+120 \right )}{20} \right ]sec=12\,sec$
Question 12 of 33

12. Question
1 points
A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
- 50 m
- 200 m
- 150 m
- Data inadequate
Correct

Incorrect

Let the length of the train be x metres and its speed by y m/sec.

Then, $\frac{x}{y}=15 \Rightarrow y=\frac{x}{15}$

$\therefore \frac{x+100}{25}=\frac{x}{15}$

15(x + 100) = 25x

15x + 1500 = 25x

1500 = 10x

x = 150 m.
Question 13 of 33

13. Question
1 points
A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
- 65 sec
- 80 sec
- 72 sec
- 82 sec
Correct

Incorrect

$speed=\left ( \frac{240}{24}\right )m/sec=10 \, \,m/sec$

$\therefore Required \, \,time=\left (\frac{240+650}{10} \right )sec=89\, \,sec$
Question 14 of 33

14. Question
1 points
A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
- 230 m
- 260 m
- 240 m
- 270 m
Correct

Incorrect

$Speed=\left ( 72\times \frac{5}{18} \right )m/sexc=20\, \,m/sec$

Time = 26 sec.

Let the length of the train be x metres.

$Then,\, \, \frac{x+250}{26}=20$

x + 250 = 520

x = 270.
Question 15 of 33

15. Question
1 points
A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
- 3.6 sec
- 36 sec
- 18 sec
- 72 sec
Correct

Speed of train relative to jogger = (45 - 9) km/hr =36 km/hr.

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

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

$\therefore Time\, \,taken\left ( \frac{360}{10} \right )sec=36 \, \,sec$

Incorrect

Speed of train relative to jogger = (45 - 9) km/hr =36 km/hr.

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

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

$\therefore Time\, \,taken\left ( \frac{360}{10} \right )sec=36 \, \,sec$
Question 16 of 33

16. Question
1 points
Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
- 9 a.m.
- 10.30 a.m.
- 10 a.m.
- 11 a.m.
Correct

Incorrect

Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x - 1) hours = 25(x - 1) km.

20x + 25(x - 1) = 110

45x = 135

x = 3.

So, they meet at 10 a.m.
Question 17 of 33

17. Question
1 points
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
- 40 sec
- 45 sec
- 42 sec
- 48 sec
Correct

Incorrect

Formula for converting from km/hr to m/s: X km/hr = $\left ( x\times \frac{5}{18} \right )m/s$

$\therefore\, \, \,speed=\left ( 45\times \frac{5}{18} \right )m/sec=\frac{25}{2}m/sec$

Total distance to be covered =(360 + 140) m=500 m.

Formula for finding Time = $\left (\frac{Distance}{Speed}\right )$

$\therefore\, \, \,Required \, \,time=\left (\frac{500\times 2}{25} \right )sec=40\, \, sec$
Question 18 of 33

18. Question
1 points
How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
- 25
- 40
- 30
- 45
Correct

Incorrect

Speed of the train relative to man = (63 - 3) km/hr= 60 km/hr

$=\left ( 60\times \frac{5}{18} \right )m/sec$

$=\left (\frac{50}{3}\right )m/sec$

Time taken to pass the man= $=\left (500\times \frac{3}{50} \right )m/sec=30\,sec$
Question 19 of 33

19. Question
1 points
Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.
- 12 sec
- 48 sec
- 24 sec
- 72 sec
Correct

Incorrect

Relative speed =(45 + 30) km/hr

$\left (75\times\frac{5}{18}\right )m/sec$

$\left (\frac{125}{6}\right )m/sec$

We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.

So, distance covered = Length of the slower train.

Therefore, Distance covered = 500 m.

Required time = $\left ( 500\times\frac{6}{125}\right )m/sec=24\,sec$
Question 20 of 33

20. Question
1 points
Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
- 36
- 48
- 45
- 49
Correct

Relative speed = (60+ 90) km/hr

$\left (150\times\frac{5}{18}\right )m/sec=\left (\frac{125}{3}\right )m/sec$

Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.

Required time = $\left (2000\times\frac{3}{125}\right )m/sec=48 \, \,sec$

Incorrect

Relative speed = (60+ 90) km/hr

$\left (150\times\frac{5}{18}\right )m/sec=\left (\frac{125}{3}\right )m/sec$

Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.

Required time = $\left (2000\times\frac{3}{125}\right )m/sec=48 \, \,sec$
Question 21 of 33

21. Question
1 points
A train 800 metres long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 minute, then the length of the tunnel (in meters) is:
- 130
- 500
- 360
- 540
Correct

Incorrect

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

Time = 1 minute = 60 seconds.

Let the length of the tunnel be x metres.

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

3(800 + x) = 3900

x = 500.
Question 22 of 33

22. Question
1 points
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
- 200 m
- 245 m
- 225 m
- 250 m
Correct

$Speed=\left ( 45\times \frac{5}{18} \right )m/sec=\left ( \frac{25}{2} \right )m/sec$

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.

Incorrect

$Speed=\left ( 45\times \frac{5}{18} \right )m/sec=\left ( \frac{25}{2} \right )m/sec$

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.
Question 23 of 33

23. Question
1 points
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
- 120 metres
- 180 metres
- 324 metres
- 150 metres
Correct

Speed= $\left(60\times\frac{5}{18}\right)$ m/s = $\frac{50}{3}$ m/s.

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

Incorrect

Speed= $\left(60\times\frac{5}{18}\right)$ m/s = $\frac{50}{3}$ m/s.

Length of the train = (Speed $\times$ Time) = $\left(\frac{50}{3}\times 9\right)$ m = 150 m
Question 24 of 33

24. Question
1 points
A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
- 65 sec
- 89 sec
- 100 sec
- 150 sec
Correct

speed = $\frac{240}{24}$ m/sec = 10 m/sec.

Required time = $\frac{240+650}{10}$ sec = 89 sec

Incorrect

speed = $\frac{240}{24}$ m/sec = 10 m/sec.

Required time = $\frac{240+650}{10}$ sec = 89 sec
Question 25 of 33

25. Question
1 points
How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 krn/hr in the direction of the moving train if the speed of the train is 63 km/hr ?
- 25
- 30
- 40
- 45
Correct

Incorrect

Speed of train relatively to man = ( 63 - 3)km/hr = 60 km /hr

= $\left(60\times\frac{5}{18}\right)$ m/sec

= $\frac{50}{3}$ m/sec

Time taken to pass the man = $\left(500\times\frac{3}{50}\right)$ sec = 30 sec.
Question 26 of 33

26. Question
1 points
Two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively. In how much time will they cross each other, if they are running in the same direction ?
- 72 sec
- 132 sec
- 192 sec
- 252 sec
Correct

Relative Speed = (45 - 40 ) Kmph = 5 kmph

= $\left(5\times\frac{5}{18}\right)$ m/sec

= $\left(\frac{5\times5}{18}\right)=\frac{25}{18}$

Time taken = $\left(350\times\frac{18}{25}\right)=14\times18$ = 252 sec

Incorrect

Relative Speed = (45 - 40 ) Kmph = 5 kmph

= $\left(5\times\frac{5}{18}\right)$ m/sec

= $\left(\frac{5\times5}{18}\right)=\frac{25}{18}$

Time taken = $\left(350\times\frac{18}{25}\right)=14\times18$ = 252 sec
Question 27 of 33

27. Question
1 points
Two, trains, one from Howrah to Delhi and the other from Delhi 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
- 2 : 3
- 4 : 3
- 6 : 7
- 9 : 16
Correct

Let us name the trains as A and B. Then,

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

Incorrect

Let us name the trains as A and B. Then,

(A's speed) : (B's speed) = $\sqrt{b}:\sqrt{a}=\sqrt{16}:\sqrt{9}$ = 4 : 3.
Question 28 of 33

28. Question
1 points
A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
- 69.5 km/h
- 79 km/h
- 70 km/h
- 79.2 km/h
Correct

Let the length of train be x meters and its speed be y m/sec

then, $\frac{x}{y}$ = 8 $\Rightarrow$ x = 8y

Now $\frac{x+264}{20}$ = y

$\Rightarrow$ 8y + 264 = 20y

$\Rightarrow$ y = 22

$\therefore$ Speed = 22 m/sec = $22\times \frac{18}{5}$ km/h = 79.2 km/h

Incorrect

Let the length of train be x meters and its speed be y m/sec

then, $\frac{x}{y}$ = 8 $\Rightarrow$ x = 8y

Now $\frac{x+264}{20}$ = y

$\Rightarrow$ 8y + 264 = 20y

$\Rightarrow$ y = 22

$\therefore$ Speed = 22 m/sec = $22\times \frac{18}{5}$ km/h = 79.2 km/h
Question 29 of 33

29. Question
1 points
Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other is
- 9
- 10
- 9.6
- 10.8
Correct

Relative speed = (60 + 40)km/h = $\left (100\times \frac{5}{18}\right)$ m/sec = $\left(\frac{250}{9}\right)m/sec$

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = $\left(300\times\frac{9}{250}\right)$ sec = $\frac{54}{5}$ sec= 10.8 sec
Relative speed = (60 + 40)km/h = $\left (100\times \frac{5}{18}\right)$ m/sec = $\left(\frac{250}{9}\right)m/sec$

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = $\left(300\times\frac{9}{250}\right)$ sec = $\frac{54}{5}$ sec= 10.8 sec" /> $ </div> <div style="display: none;" class="wpProQuiz_incorrect"> Incorrect Relative speed = (60 + 40)km/h = <img src='https://questionpaper.org/wp-content/plugins/latex/cache/tex_d95a3b3eac8d51f332116754f2a5e41d.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\left (100\times \frac{5}{18}\right)" /><script type='math/tex'>\left (100\times \frac{5}{18}\right)$ m/sec = $\left(\frac{250}{9}\right)m/sec$

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = $\left(300\times\frac{9}{250}\right)$ sec = $\frac{54}{5}$ sec= 10.8 sec
Question 30 of 33

30. Question
1 points
A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
- 66 km/hr
- 78 km/hr
- 72 km/hr
- 81 km/hr
Correct

4.5km/h = $\left ( 4.5\times \frac{5}{18} \right )$ m/sec = $\frac{5}{4}$ m/sec = 1.25 m/sec

5.4km/h = $\left ( 5.4\times \frac{5}{18} \right )$ m/sec = $\frac{3}{2}$ m/sec = $1.5 m/sec Let the speed of the train be x m/sec. Then, (x - 1.25)�$ \times $8.4 = (x - 1.5)�$ \times $8.5 8.4x - 10.5 = 8.5x - 12.75 0.1x = 2.25 x = 22.5 Speed of the train =$ \left ( 22.5\times\frac{18}{5}\right )
4.5km/h = $\left ( 4.5\times \frac{5}{18} \right )$ m/sec = $\frac{5}{4}$ m/sec = 1.25� m/sec

5.4km/h = $\left ( 5.4\times \frac{5}{18} \right )$ � m/sec = $\frac{3}{2}$ m/sec = $1.5 m/sec Let the speed of the train be x m/sec. Then, (x - 1.25)?$ \times $8.4 = (x - 1.5)?$ \times $8.5 8.4x - 10.5 = 8.5x - 12.75 0.1x = 2.25 x = 22.5 Speed of the train =$ \left ( 22.5\times\frac{18}{5}\right )" /> $km/h = 81 km/h </div> <div style="display: none;" class="wpProQuiz_incorrect"> Incorrect 4.5km/h = <img src='https://questionpaper.org/wp-content/plugins/latex/cache/tex_8e92a3a46673e24db4e752d30798f4c7.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\left ( 4.5\times \frac{5}{18} \right )" /><script type='math/tex'>\left ( 4.5\times \frac{5}{18} \right )$ m/sec = $\frac{5}{4}$ m/sec = 1.25� m/sec

5.4km/h = $\left ( 5.4\times \frac{5}{18} \right )$ � m/sec = $\frac{3}{2}$ m/sec = $1.5 m/sec Let the speed of the train be x m/sec. Then, (x - 1.25)?$ \times $8.4 = (x - 1.5)?$ \times $8.5 8.4x - 10.5 = 8.5x - 12.75 0.1x = 2.25 x = 22.5 Speed of the train =$ \left ( 22.5\times\frac{18}{5}\right ) km/h = 81 km/h
Question 31 of 33

31. Question
1 points
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
- 1 : 3
- 3 : 4
- 3 : 2
- None of these
Correct

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x metres,

and length of the second train = 17 y metres.

$\therefore \frac{27x+17y}{x+y}=23$

27x + 17y = 23x + 23y

4x = 6y

$\Rightarrow \frac{x}{y}=\frac{3}{2}$

Incorrect

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x metres,

and length of the second train = 17 y metres.

$\therefore \frac{27x+17y}{x+y}=23$

27x + 17y = 23x + 23y

4x = 6y

$\Rightarrow \frac{x}{y}=\frac{3}{2}$
Question 32 of 33

32. Question
1 points
A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
- 5 sec
- 7 sec
- 6 sec
- 10 sec
Correct

Incorrect

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

$=\left ( 66\times \frac{5}{18} \right )m/sec$

$=\left ( \frac{55}{3} \right )m/sec$

$\therefore Time\,taken\,to \,pass\,the\,man=\left (110\times \frac{3}{55} \right )sec=6 \, \,sec$
Question 33 of 33

33. Question
1 points
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:
- 2 : 3
- 6 : 7
- 4 : 3
- 9 : 16
Correct

Incorrect

Let us name the trains as A and B. Then,

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