EXAMVEDA

1.
Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a
man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8
minutes?
 Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20+x) kmph.
Or, 20*10/60 = 8/60(20+x)
Or, 200 = 160+8x
Or, 8x = 40
Hence, x = 5kmph.

Detailed Explanation:
A _____________M_______________B
A = Bus Terminal.

B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e.
Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10
minutes as well.

Distance Covered by Bus in 10 min = AB = (20/60) * 10 = 10/3 km.

Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes.
Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction
of each other. Let Speed of the man = x kmph.

Relative speed = 20 +x

To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed.
And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with
relative speed (20 +x) kmph = distance covered by bus in 10 mintes with speed 20 kmph.
3.
Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively.
How many kilometers from Delhi will the two trains be together:
Difference in time of departure between two trains = 45 min. = 45/60 hour = 3/4 hour.
Let the distance be x km from Delhi where the two trains will be together.
Time taken to cover x km with speed 136 kmph be t hour
and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be
(t+3/4) 
= ((4t+3)/4);

Now,
100*(4t+3)/4 = 136t;
Or, 25(4t+3) = 136t;
Or, 100t+75 = 136t;
Or, 36t = 75;
Or, t = 75/36 = 2.08 hours;
Then, distance x km = 136*2.084 = 283.33 km.
5.
A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60%
of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:
Let the total distance be 100 km.
Average speed = total distance covered/ time taken;
= 100/[(30/20)+(60/40)+(10/10)];
= 100/[(3/2)+(3/2)+(1)];
= 100/[(3+3+2)/2]
= (100*2)/8
= 25 kmph.
9.
The speed of A and B are in the ratio 3:4. A takes 20 minutes more than B to reach a destination. Time in
which A reach the destination?
Ratio of speed = 3:4;
Ratio of time taken = 4:3 (As Speed ∝ (1/Time), When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.

Then,
4x-3x = 20/60;
Or, x = 1/3;
Hence, time taken by A = 4x 
hours = 4*1/3 = 1(1/3) hours.
14.
A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 72 minutes, will be
covered by A in:
Ratio of the speed of A, B and C = 6:3:1
Then, ratio of time taken;
= 1/6:1/3:1 = 1:2:6;
Hence, time taken by A = 72/6 = 12 minutes.
22.
Two trains started at the same time, one from A to B and other from B to A. if they arrived at B and A
respectively in 4 hours and 9 hours after they passed each other, the ratio of the speeds of the two trains was:
A →____________________← B 
Ratio of the speed of trains is given by;
(Speed of train A/ Speed train of B) = √b/√a.
= √9/√4.
= 3/2
= 3:2.
 23.
Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will
cross each other in:
 Let length of each train be x meter.
Then, speed of 1st train = x/18 m/sec;
Speed of 2nd train = x/12 m/sec;

Now,
When both trains cross each other, time taken;
= [2x/(x/{18)+(x/12)}]
= 2x/[(2x+3x)/36] = 2x*36/5x
= 72/5
= 14.4 seconds.
25.
Two trains 105 meters and 90 meters long, run at the speeds of 45 kmph and 72 kmph respectively, in
opposite directions on parallel tracks. The time which they take to cross each other, is:
 Length of the 1st train = 105 m;
Length of the 2nd train = 90 m.
Relative speed of the trains,
= 45+72 = 117kmph
= 117*5/18 = 32.5 m/sec;
Time taken to cross each other,
= (Length of 1st train + length of 2nd train) /relative speed of the trains.
Time taken = 195/32.5 = 6 secs.
26.
A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10
seconds and 11 seconds respectively. The speed of the train is
1st method:
Let the speed of the train be S. And length of the train be x.
When a train crosses a man, its travels its own distance.

According to question;
x/[(S-3)*(5/18)] = 10;
or, 18x = 50*S-150; . . . . . . . . (i)
And;
x/[(x-5)*(5/18)] = 11; 
18x = 55*S-275 . . . . . . . . . .. . . (ii)
Equating equation (i) and (ii), we get,
50*S-150 = 55*S-275;
Or, 5*S = 125;
Or, S = 25 kmph.

2nd method:
In the both cases train has to travel its own length that means, distance is constant.
So, we have,
When distance is constant, speed is inversely proportional to time.
i.e. speed α 1/time.
In such case, the following ratios will be valid:&l
27.
A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The
length of the train is:
1st Method:
Let length of the train be x m and speed of the train is s kmph.
Speed, s = (x+800)/100 ..... (i)
Speed, s = (x+400)/60; ..... (ii)
Equating equation (i) and (ii), we get,
Or, (x+800)/100 = (x+400)/60;
Or, 5x+200 = 3x+2400;
Or, 2x = 400;
Or, x = 200 m.

2nd Method:
As in both cases, the speed of the train is constant, and then we have;
Time α distance
100/60 = (x+800)/(x+400);
Or, x = 200 m.
28.
A train travelling with a speed of 60 kmph catches another train travelling in the same direction and then
leaves it 120 m behind in 18 seconds. The speed of the second train is
Let speed of the 2nd train is S m/sec.
And,
60 kmph = (60*5)/18 = 50/3 m/sec.
As trains are traveling in same distance, Then Relative distance, 
(50/3)-S = 120/18;
Or, (50-3S)/3 = 20/3;
Or, 50-3S = 20;
S = 10 m/sec. 
Or, Speed of the 2nd train = 10*18/5 = 36 kmph.
36.
A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers
away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.
By increasing the speed by 33.33%, it would be able to reduce the time taken for traveling by 25%.
But since this is able to overcome the time delay of 30 minutes, 30 minutes must be equivalent to
25% of the time originally taken.
Hence, the original time must have been 2 hours and the original speed would be 750 kmph. 
Hence, the new speed would be 1000 kmph.
37.
A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in the same
direction. The pedestrian could see the car for 6 minutes and it was visible to him up to distance of 0.6 km.
what was speed of the car?
 In 6 minutes, the car goes ahead by 0.6 km.
Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph.
Hence, Net speed of the car is 8 kmph.
40.
Two rifles are fired from the same place at a difference of 11 minutes 45 seconds. But a man who is coming
towards the place in a train hears the second sound after 11 minutes. Find the speed of train. (speed of sound
= 330 m/s)
Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min)
= 45 sec.
Let the speed of train be x kmph.
Now,
Or, x*11/60 = (45/3600)*330*18/5; 
Or, x = 81 kmph.
41.
Two horses started simultaneously towards each other and meet each other 3 hr 20 mins later. How much
time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of
the second 5 hours later than the second arrived at the point of departure of the first?
Since, the two horses meet after 200 minutes; they cover 0.5% of the distance per minute (combined)
or 30% distance per hour.
This condition is satisfied only if the slower rider takes 10 hours (thereby covering 10% per hour) and
the faster rider takes 5 hours.
42.
Two ports A and B are 300 km apart. Two ships leave A for B such that the second leaves 8 hours after the
first. The ships arrive at B simultaneously. Find the time the slower ship spent on the trip if the speed of one
of them is 10 km/h higher than that of the other.
A__________300 km_______B
Let speed of the 1st ship is s kmph and speed of 2nd ship is (s+10) kmph.
Let 2nd ship takes time t to cover the distance 300 km then 1stship takes (t+8) hours.

As distance is constant, we have,

s/(s+10) = t/(t+8);
Or, st+8s = st+10t;
Or, 8s = 10t;

If the slower ship took 20 hours (option d) the faster ship would take 12 hours and their respective
speeds would be 15 and 25 kmph.
 45.
A ship leaves on a long voyage. When it is 18 miles from shore, a sea plane, whose speed is ten times that of
the ship, is sent to deliver mail. How far from the shore does the sea plane catch up with the ship?
A _______18miles_____B ___x(let)___C
Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h.
Let time t is taken to cover the distance x from point B to point C by ship.
Hence, in time t plane will cover (18+x) miles.

Now,
x/1 = (18+x)/10;
Or, 10x = 18+x;
Or, x = 2 miles.
Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time.

Short method:
Relative speed = (10-1) = 9 miles/h. 
hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.
49.
Two trains start from the same point simultaneously and in the same direction. The first train travels at 40 km
/h, and the speed of the second train is 25% more than the speed of first train. Thirty minutes later, a third
train starts from same point and in the same direction. It over takes the second train 90 minutes later than it
overtook the first train. What is the speed of the third train?
A _______ B ________ C 
Let both the trains start from point A.
At point B third train overtook the first train.
To overtake the first train by third train, Third train needs to cover,
Distance covered by first train in 1(1/2) h = distance covered by third train in 1 h. 
[as third train has stared 30 minutes later]
In this situation distance is constant,
then from, 
S*T = D; we get, S α 1/T;
Now, (t+1/2)/t = S/40;
(2t+1/t) = S/40; . . . . . . (1)
From equation (1), 
it is clear that time is in the ratio 3:2 then speed will be in 2:3 ratios. 
Hence, Speed of the Third train will be 60 km/h.
 55.
Due to the technical snag in the signal system two trains start approaching each other on the same track from
two different stations, 240 km away each other. When the train starts a bird also starts moving to and fro
between the two trains at 60 kmph touching each train each time. The bird initially sitting on the top of the
engine of one of the trains and it moves so till these trains collide. If these trains collide one and half hour
after start, then how many kilometers bird travels till the time of collision of trains?
 Time taken to collide the trains,
= one and half hour = 3/2 h.
So, in 3/2 h bird travels,
= (3*60)/2 
= 90km.
60.
A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. 5/12 of the distance AB
measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train
catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?
Train(T)__________ A_____5k____CAT__________B

T<-------x--------------><----12k-------------------------->

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from
A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
=> v /u = x/5k = (x+12k)/7k 
=> 7x = 5(x+12k)
x/k = 30/1
Thus, 
=> u/v = 30/5 = 6/1.
61.
The driver of an ambulance sees a school bus 40 m ahead of him after 20 seconds, the school bus is 60 meter
behind. If the speed of the ambulance is 30 km/h, what is the speed of the school bus?
Relative Speed,
= (Total distance)/total time 
= (60+40) /20 
= 5 m/s = (5*18)/5 = 18 kmph
Relative Speed = (speed of ambulance - speed of school bus)
Speed of school bus = speed of ambulance - relative speed.
= 30-18 = 12 kmph.
62.
A minibus takes 6 hour less to cover 1680 km distance, if its speed is increased by 14 kmph ? What is the
usual time of the minibus ?
Let 
Usual Speed = X kmph.
Now, 
Speed of bus after increasing the speed = (X +14)kmph. -------------- (1)
A ___________________1680km __________________B
In first case, 
Time taken to covered the distance 1680 km = 1680/X ----------- (2) 
In Second Case, 
Time Taken to covered the distance 1680 km = 1680/(X + 14).
Time difference = 6 Hours.
So,

[(1680/X) - 1680/(X + 14)] = 6.

[1680X +23520 - 1680X]/{X *(X+14)} = 6
23520 = 6X2 +84X.
6X2 +84X -23520 = 0.
X 2+ 14X - 3920 = 0
X2 + 70X - 56X - 3920 = 0
X (X + 70) - 56 (X + 70) = 0

X = -70, 56.
X is original speed so it cannot be negative.
64.
A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio
of the two speed is :
 Let the original speed be S1 and time t1 and distance be D.
Now,
(D/2) /2t1 =S2
S2 = D/4t1 and S1 = D /t1
Thus,
S1 /S2 = 4/1 = 4:1.
A is twice fast as B and B is thrice fast as C. The journey covered by C in 78 minutes will be covered
by A in :
The ratio of speeds of A, B, C = 6:3:1.
The ratio of time taken by A, B, C = (1/6):(1/3):(1) To simplify it, we will multiply it by LCM of ratio of
speeds given. Hence, the ratio of time taken by A, B, C = 1:2:6.
[Speed is inversely proportional to time, means if speed increase time decreases. So, ratio of time
would be reciprocal of the ratio of speed given. ] Time taken by C to covered given distance = 78 =
6*13. The ratio of time of A and C = 1 :6 Thus, time taken by A = 13 min.

TrainINDIABIX
1.  A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of
the train?

Speed 60 x 5 = 50
 18 3
=
m/sec m/sec.
Length of the train = (Speed x Time).
50
 Length of the train = x9
m = 150 m.

2.  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:

12
Speed of the train relative to man 5
= m/sec
10
   25
= 2 m/sec.
   25 18
x
= 2 5 km/hr
   = 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.

3.  The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross
in 30 seconds, is:

Speed 5 25
45 x =
= 18 m/sec 2 m/sec.
Time = 30 sec.
Let the length of bridge be x metres.
130
Then, + x = 25
30 2
 2(130 + x) = 750
 x = 245 m.

4.  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:

Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
27x +
17y = 23
x+ y
 27x + 17y = 23x + 23y
 4x = 6y
 x 3
= .
y 2

5.  A train passes a station platform in 36 seconds and a man standing on the platform in 20
seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

Speed 5
54 x
= 18 m/sec = 15 m/sec.
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
x +
Then, 300 = 15
36
 x + 300 = 540
 x = 240 m.

6.  A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650
m long?

Speed 240
= 24 m/sec = 10 m/sec.
 Requir 240 + 650
ed time = 10 sec = 89 sec.
7.  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:

Let the length of each train be x metres.

Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
   10 5
= x 18 m/sec
   25
= 9 m/sec
2
x 25
=
3 9
6
 2x = 100
 x = 50.

8.  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?
Formula for converting from km/hr to m/s:   X km/hr 5
X x m/s.
= 18
45 5 25
Therefore, Speed = = m/sec.
x 18 m/sec 2
Total distance to be covered = (360 + 140) m = 500 m.
Distanc
Formula for finding e
Time =
Speed
500 x 2
 Required time = = 40 sec.
sec

9.  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:
Relative speed = (60+ 90) km/hr
   5
150 x
= 18 m/sec
12
   5
= m/sec.
3
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
3
Required time
2000 x 12
= sec = 48 sec.
5

10. 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?

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

   5
36 x
= 18 m/sec
   = 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
360
 Time taken = = 36 sec.
10 sec

11. 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?

Relative speed = (120 + 80) km/hr

   5
200 x
= 18 m/sec
 50
   0
= m/sec.
9
Let the length of the other train be x metres.
Then x + 270 500
=
, 9 9
 x + 270 = 500
 x = 230.

12.  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?

Speed 5
72 x = 20 m/sec.
= 18 m/sec
Time = 26 sec.
Let the length of the train be x metres.
x +
Then, 250 = 20
26
 x + 250 = 520
 x = 270.

13. 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:

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.
(100 + 100) =
8 3x
 24x = 200
2
 x = 5 .
3
So, speed of the faster train 50
m/sec
= 3
   50 18
x
= 3 5 km/hr
   = 60 km/hr.

14.  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:
Relative speed = (60 + 40) km/hr 5 250
100 x =
= 18 m/sec 9 m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Require 9 54
300 x = sec = 10.8 sec.
d time = 250 sec 5
15.  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?

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

   66 5
= x 18 m/sec
   55
= 3 m/sec.
3
 Time taken to pass the man = 110 x
55 sec = 6 sec.

16. 
A train travelling at a speed of 75 mph enters a tunnel 3  miles long. The train
is   mile long. How long does it take for the train to pass through the tunnel
from the moment the front enters to the moment the rear emerges?

Total distance 7 1
= + miles
covered 2 4

15
= miles.
4

15
 Time taken = hrs
4 x 75

1
= hrs
20

1
= x 60 min.
20

= 3 min.

A train 800 metres long is running at a speed of 78 km/hr. If it crosses a tunnel in

17. 
1 minute, then the length of the tunnel (in meters) is:

Speed 5 65
78 x m/sec = m/sec.
= 18 3
Time = 1 minute = 60 seconds.
Let the length of the tunnel be x metres.
 800
Then, + x = 65
60 3
 3(800 + x) = 3900
 x = 500.

18.  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?

Speed 300 50
m/sec = m/sec.
= 18 3
Let the length of the platform be x metres.
x +
Then, 300 = 50
39 3
 3(x + 300) = 1950
 x = 350 m.

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
Let the length of the train be x metres and its speed by y m/sec.
x x
Then, = 15           y = .
y 15
x +
100 = x
25 15
 15(x + 100) = 25x
 15x + 1500 = 25x
 1500 = 10x
 x = 150 m.

20. 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?

Let the length of the train be x metres and its speed by y m/sec.
x
Then, = 8           x = 8y
y
x +
Now 264 = 
, y
20
 8y + 264 = 20y
 y = 22.
 Speed = 22 18
22 x km/hr = 79.2 km/hr.
m/sec = 5
21.  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?
Speed of the train relative
= (63 - 3) km/hr
to man

= 60 km/hr

5
= 60 x m/sec
18

50
= m/sec.
3

 Time taken to pass 3

= 500 x sec
the man 50

= 30 sec.

Two goods train each 500 m long, are

running in opposite directions on parallel
tracks. Their speeds are 45 km/hr and 30
22. 
km/hr respectively. Find the time taken by
the slower train to pass the driver of the
faster one.

Relative speed = = (45 + 30) km/hr

5
= 75 x m/sec
18

12
= 5 m/sec.
6

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.
6
 Required time = 500 x = 24 sec.
125

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:
Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
(120 + 120)
So, 2x =
12
  2x = 20
 x = 10.
 Spe 18
ed of
each
train = 10 x km/hr = 36 km/hr.
10
m/sec
=
24.  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?

Speed of the first train 120

m/sec = 12 m/sec.
= 10
Speed of the second train 120
m/sec = 8 m/sec.
= 15
Relative speed = (12 + 8) = 20 m/sec.
(120 +
 Required time = 120) sec = 12 sec.
20

25. 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:

Let the speed of the second train be x km/hr.

Relative
= (x + 50) km/hr
speed

(x + 50) 5
= m/sec
x 18

250 +
= 5x m/sec.
18

Distance covered = (108 + 112) = 220 m.

220
250 +
=6
5x
18
 250 + 5x = 660
 x = 82 km/hr.

26.  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?
Relative speed = (40 - 20) km/hr 20 x 5 m/sec = 50 m/sec.
= 18 9

50 250 7
 Length of faster train = x5 m= m = 27 m.
9 9 9

27.  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:
5 5
2 kmph = 2x m/sec = m/sec.
18 9
4 kmph 5 10
4x m/sec = m/sec.
= 18 9
Let the length of the train be x metres and its speed by y m/sec.
x x
Then, 5 = 9 and 10 = 10.
y - y -
9 9
 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.

28.  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?
4.5 km/hr 5 5
4.5 x m/sec = m/sec = 1.25 m/sec, and
= 18 4
5.4 km/hr 5 3
5.4 x m/sec = m/sec = 1.5 m/sec.
= 18 2
Let the speed of the train be x m/sec.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
 8.4x - 10.5 = 8.5x - 12.75
 0.1x = 2.25
 x = 22.5
 Spe 18
ed of
22.5 x km/hr = 81 km/hr.
the
train =
29.  A train travelling at 48 kmph completely crosses another train having half its length and
travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in
45 seconds. The length of the platform is

Let the length of the first train be x metres.

Then, the length of the second train x
metres.
is 2
Relative speed = (48 + 42) kmph 90 5
m/sec = 25 m/sec.
= x 18
[x + = 12 3 = 300     or     x = 200.
 (x/2)] x
or
25 2
 Length of first train = 200 m.
Let the length of platform be y metres.
Speed of the first train 5 40
48 x m/sec = m/sec.
= 18 3
3
 (200 + y) x 4 = 45
0
 600 + 3y = 1800
 y = 400 m.

30. 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?

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

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

(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.

Sawaal
A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite
direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?

Relative speed =   m / sec =   kmph = 112 kmph.

Speed of goods train = (112 - 50) kmph = 62 kmph.

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 ?
Let the length of the train be x metres and its speed by y m/sec.
Then,
Now,   
8y + 264 = 20y
 y = 22.
Speed = 22 m/sec = km/hr = 79.2 km/hr.s
A train of length 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is
running at 6 kmph in the direction opposite to that in which the train is going?
Distance = 110 m

Relative speed = 60 + 6 = 66 kmph (Since both the train and the man are in moving in opposite
direction)

=   m/sec =   m/sec

Time taken to pass the man =   = 6 s

Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the
length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling
in opposite direction?

Speed of train 1 =   = 12 m/sec

Speed of train 2 =   = 8 m/sec

if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec

distance covered = 120 + 120 = 240 m

time = distance/speed = 240/20 = 12 sec

A train travelling at a speed of 75 mph enters a tunnel  miles long. The train is  mile long. How long
does it take for the train to pass through the tunnel from the moment the front enters to the moment
the rear emerges?

Total distance covered = miles

= miles 

Time taken =  hrs

=  hrs

=  min

= 3 min
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:
Relative speed = (60 + 40) km/hr =[ 100 x ( 5 / 18 ) ]m/sec = ( 250 /9 ) m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [ 300 x ( 9/250 ) ] sec = ( 54/ 5 )sec = 10.8 sec.
Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and
travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed
of 25 kmph. At what time will they meet?
Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3

Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
Two trains started at the same time, one from A to B and the other from B to A . If they arrived at B
and A respectively 4 hours and 9 hours after they passed each other the ratio of the speeds of the
two trains was 

If two trains (or bodies) start at the same time from points A and B towards each other and after
crossing they take a and b sec in reaching B and A respectively, then: (A's speed) : (B's speed) =
(b : a)

Therefore, Ratio of the speeds of two trains =   = 3 : 2

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 ?

Relative speed = (120 + 80) km/hr

= =

Let the length of the other train be x metres.

Then, 
  x + 270 = 500

 x = 230.

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 ?

Speed =  m/sec =  m/sec.

Let the length of the platform be x metres.

Then,  =     3(x+300) = 1950    x = 350m

Two trains of equal length , running in opposite directions , pass a pole in 18 and 12 seconds. The
trains will cross each other in 

Let the length of the train be L metres

Speed of first train =   m/hour

Speed of secxond train =   m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken =   sec

                 =14.4 sec

Two trains are running in opposite directions in the same speed. The length of each train is 120
meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v + v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s 

= 10 × 36/10 km/hr = 36 km/hr

How many seconds will a 500 meter long train moving with a speed of 63 km/hr  take to cross a man
walking with a speed of 3 km/hr in the direction of the train ?
Distance = 500 m

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

=   m/s =  m/s

Time taken to pass the man = distance/speed =   sec = 30 sec

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?

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

Time taken to pass the man =   = 6 sec.

A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's
speed against the current is 
A train running at the speed of 40 km/hr crosses a signal pole in 9 seconds. Find the length of the
train ?
two trains 120m and 80m in length are running in opposite directions with velocities 42km/hour and
30 km/hour then at what time will they completely cross each other?

Suppose the one train is at rest then the relative velocity of other train with respect to this train will
be= sum of velocities of the two trains that is 42+30=72km/hr or 20m/s .Distance that it will have to
cover to cross the other train completely=sum of the lengths of the two trains that is 120+80=200m.
Time required=distance/speed(relative speed) =200/20=10sec. 
120m train running with 126km/h and crosses the 150m long train in 27 sec in same direction then
what will be the speed of second train?

Total distance covered by first train to cross the second train = 120+150 = 270 m
Let velocity of second train = x km/h
Relative velocity of trains= (126-x) km/h = (126-x)5/18 m/s
v=d/t
(126-x)5/18 = 270/27
x = 90
So, speed of second train will be 90 km/h.

A person meets a train at a railway station coming daily at a particular time. One day he is late by 25
minutes, and he meets the train 5 km before the station. If his speed is 12 kmph, what is the speed of
the train?

ALGEBRA.COM
A private plane and a commercial plane take off from an airport at the same time for a city 720 km away. The
speed of the private plane is 180km/h less than that of a commercial plane. If the commercial plane arrives 2
hours ahead of the private plane , find each planes rate 
Let the rate of the commercial plane be x. 
Then the rate of the private plane is x - 180. 

Using time = distance / rate, the time it takes each plane to travel is 

720/x for the commercial plane 

720/(x - 180) for the private plane 

If the commercial plane arrives two hours ahead of the private plane, then 

720/(x - 180) - 720/x = 2 

multiply both sides by x(x - 180) 

720x - 720(x - 180) = 2x(x - 180) 

720x - 720x + 720*180 = 2x^2 - 360x 

2x^2 - 360x - 129600 = 0 

x^2 - 180x - 64800 = 0 

x = (180 ± √(180^2 + 4*64800))/2 

x = (180 ± 540)/2 

x = 360 or -180 

we reject x = -180, so we get x = 360 

the rate of the commercial plane is 360 mph 

the rate of the private plane is 180 mph 

check 
the private plane reaches the city in 720/180 = 4 hours 
the commercial plane reaches the city in 720/360 = 2 hours 
So the commercial plane arrives two hours ahead of the private plane.
The speed limit on a highway is 60 miles per hour. A vehicle travels for 2.2 hours on the highway and covers
165 miles. By how many miles per hour is the driver over/under the speed limit?
d = r*t
r = d/t
==================
To find mi/hr, divide the miles by the hours

A person drives 390 miles on a stretch

of road. Half the distance is driven traveling 5
miles per hour below the speed limit, and half the
distance is driven traveling 5 miles per hour above
the speed limit. If the time spent traveling at the slower
speed exceeds the time spent traveling at the faster
speed by 24 minutes, find the speed limit. 
Let the speed limit be x mph 
First half speed = (x-5) mph
Second half speed = (x+5) mph 
Half distance = 390/2 = 195 miles 
24 minutes = 24/60 = 2/5 hours 
time = distance/speed 
First half time = 195/(x-5) 
Seond half time = 195/(x+5) 

 
LCD = 5(x+5)(x-5) 
Multiply equation by the LCD
 
975x+4875-975x+4875 = 2(x^2-25) 
9750 = 2x^2-50
2x^2= 9800
/2
x^2= 4900
take the square root
x= 70 mph 
Speed limit is 70 mph 

An Amtrak train took 10 hours to travel the same distance that the Metra train traveled in 15
hours.
Find the speed of each train if the speed of the Amtrak train was 30 mph faster the Metra
train. 
:
let s = speed of the Metra train
" the speed of the Amtrak train was 30 mph faster the Metra train." therefore
(s+30) = speed of Amtrak train
:
Write distance equation, dist = time * speed
" An Amtrak train took 10 hours to travel the same distance that the Metra train traveled in 15
hours." therefore:
15s = 10(s + 30)
15s = 10s + 300
15s - 10s =300
5s = 300
s = 300/5
s = 60 mph is the Metra train speed
then, obviously
60 + 30 = 909 mph is the Amtrak speed
;
;
Confirm this find the actual time for each, they should be equal
10 * 90 = 900 mi
15 * 60 = 900 mi

a plane flies the first half of a 5600 km flight into the wind in 3.5 hours. the return trip, with the same wind,
takes 2.5 hours. find the speed of the wind and the speed of the plane in still air 
Plane speed= x mph
wind speed= y mph
with wind speed= (x+y) 
against wind speed (x-y) 
d= 2800 =
2800 / (x+y) 2.5
divide by 2.5 = 1
1120 / (x+y) 
(x+y) = 1120 ............1
d= = 2800 against wind
2800 /(x-y)= 3.5
divide by 3.5
800 /(x-y) = 1
x - y = 800 .............2
add up (1) & (2)
2 x =1920
/2
x= 960 mph speed of Plane in still air
plug value of x in (1)
we get y= 160 mph speed of wind

Jacob started biking to the coffee shop traveling 9 mph, after some time the bike got a flat so Jacob walked the
rest of the way, traveling 2 mph. If the total trip to the coffee shop took 9 hours and it was 60 miles away, how
long did Jacob travel at each speed? 
let b = time spent biking
then
(9-b) = time spent walking
:
Write a distance equation, dist = speed * time
bike dist + walk dist = 60 mi
9b + 2(9-b) = 60
9b + 18 - 2b = 60
9b - 2b = 60 - 18
7b = 42
b = 42/7
b = 6 hrs spend biking
then
9 - 6 = 3 hrs spent walking
:
:
:
Confirm this, find that actual dist for each
9(6) = 54 mi
2(3) = 6 mi

Julie and George can row at the same rate in stillwater. They leave Carsten at the same time, Julie is going
upstream and George going downstream. Julie rows for 2 hours and arrives in Colton, three miles from
Karsten. George rown for 3 hours and arrives at Burnburg, 13.5 miles from Karsten. Find the rate of the current
and the rate each rows in stillwater 
Let r be the rate they can row in still water, and c be the rate of the current. Then:
2(r-c)=3;
and 3(r+c)=13.5. So:
2r-2c=3
3r+3c=13.5
6r-6c=9
6r+6c=27
12r=36
r=3
c=3/2
They both row at 3 mph; and the current is 1.5 mph.

Julie and Eric row their boat (at a constant speed) 45 miles downstream for 5 hours, helped by the current.
Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the current. 
Let the current be 
Let the rate of the boat with no current be 
With the current:
Against the current:

-------------------
Given:

Rewriting these,

Multiply the 1st by   and the 2nd by 

Add these equations

Divide both sides by 

Substitute this back into a given equation

The rate of the current is 2 mi/hr

check:

------------

OK

it is 280 miles from Memphis to Birmingham. An expres train left Memphis traveling towards Birmingham at
70 mi/h. Two hours earlier a freight train left Birmingham traveling towards Memphis on a parallel track. The
speed of this train was 50 mi/h. How long was it before the two trains meet? 
SPEED TIME DISTANCE
M to B 70 t 70t
B to M 50 t+2 50(t+2)
Total 280

 
The two trains meet 1.5 hours after the Memphis to Birmingham train leaves.

Two planes leave O'Hare Airport in Chicago at the same time. One flies directly east at 540 mph. The other flies
directly west at 720 mph. How long will it take them to be 1890 miles apart? 
Using SPEED(s)=DISTANCE(D)/time(t) 

...Make Distance the subject 

Then,let D1=st....distance eastward 

And D2=st 

..So substitute speed 

..D1=540t and D2=720t 

..because they are 1890 apart,so we say D1+D2=1890 

Hence; 540t+720t=1890 

..1260t=1890 

..T=1890/1260 

..T=1.5....meaning...one and a half hour. 

Two cars leave town at the same time going in opposite directions. one travels at 35 km/h and the other at 55
km/h. in how many hours will they be 200 km apart? 
let x be the time taken for 200 km apeart.
in x hrs car "A" will go x*35 km (distance =speed * time)
in x hrs car "B" will go x*55 km 
togather both will go 35x+55x = 200
90x = 200
x=200/90 hts
x= 2 hrs and 2/9 hrs
= 2 hrs and 2/9*60 mins
= 2 hrs and 13 mins
Two trains, one 500 meters long and the other 1000 meters long, are traveling in opposite directions on
parallel tracks. The first train is traveling at 100km/h and the other at 80km/h. How long will it take for the
trains to pass each other from the moment the locomotives are at the same point? 
a bullet train which is 40km/hr faster than the existing train will replace the existing train on an 850km
journey. If the bullet train does the journey 3hours faster than the existing train, what are their speeds? 
Let V km/hr be the speed of the bullet train and so the speed of the 
existing train is V - 40. 
The time for the the bullet train to finish the whole journey is
850/V, while it takes 850/(V-40) hr for the existing train for the
journey. Since we know the time difference is 3 hrs, we have 
3 + 850/V = 850/(V-40), 
Cancel the denominator by multiply V(V-40):
3V(V-40) + 850(V-40) = 850 V,
Simplify: 3 V^2 - 120 V - 34000 = 0,

By quadratic formula,we get V = 20+ 200/3 SQRT(94) = 84.64 mile/hr

(the speed of the bullet train)
or 20 - 200/3 SQRT(94) [negative,invalid]
and so the speed of of the existing train is 44.64 miles/hr 

If two cars leave the same location at the same time, And car one is traveling 20 MPH faster than car two in
opposite directions, and they are 168 miles apart 1 1/2 hours later...how fast is each one going?
Assume the speed of car one is x mph, 
then the speed of car two is x-2 mph. 
When they travel in the opposite direction,
distance = sum of two speed * Time, 
So, 168 = (x+x-2)* (3/2) = 3x -3,
or 3x = 171, x = 57 and x-2 = 55. 
Answer: speed of car 1 is 57 MPH,
speed of car 2 is 55 MPH

Let x be the slower car

x = slower car
x+20 = faster car
168/1.5 = 112 = how much they are apart in one hour
x+x+20 = 112
2x = 92
x = 46
So the speed of the slower car is 46mph, the speed of the faster car is 66mph

A deer is standing 80m in from the west end of a tunnel. The deer sees a train approaching from the west at a
constant speed ten times the speed the deer can run. The deer reacts by running toward the train and clears
the exit when the train is 20m from the tunnel. If the deer ran in the opposite direction it would be hit by the
train at the Eastern entrance. 
How long is the tunnel? 
80m ***** L - 80 m
Train <----Deer----------------->
Speed of the deer V, speed of the train 10V.
When the deer running toward the train, the time T it takes is 80/V = T.
and in the same time the train travels 10 V T = 10 V * 80/V = 800 m.
(where it is 20m from the tunnel)
Hence,the diatance from the train to the west entrance of the tunnel is 
800+20 = 820 m.
And, we see the dstance between the train and the deer is 820+80 = 900 m. 
Assume the lenght of the tunnel is L meters.
When the deer running toward the east, it takes time = 900/(10V - V)= 
100/V for the train to reach 
(hit) the deer (at the Eastern end). 
Hence, L-80 = V (100/V) = 100, and we obtain the length of the tunnel is 
180 m. 

A postal truck leaves its station and heads for Chicago, averaging 40mph. An error in the mailing schedule is
spotted and 24 min. after the truck leaves, a car is sent to overtake the truck. If the car averages 50mph, how
long will it take to catch the postal truck? 
If the give two speeds are v1(< v2) with distance D,then
the time for the 2nd vehicle to overtake the 1st vehicle is
T = D/(v2 - v1). 
Now,D = 40* 24/60 = 40* 2/5 = 16 miles, 
Hence, T = 16/(50-40) = 1.6 hours