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31. A train takes 18 seconds to pass a 162 metres long station and 15 seconds to pass another 120 metres long station. Find the length of the train and its speed in km/hr.
  A.  Length 95 metres; Speed 55 km/hr
  B.  Length 90 metres; Speed 50.4 km/hr
  C.  Length 100 metres; Speed 51.4km/hr
  D.  Length 92 metres; Speed 54.4 km/hr
     
   
View Answer

Let the length of the train = y metre
length of station = 162
Total distance = (y + 162) metre
time to cross station = 18 sec
(y + 162) / 18
=
(y + 120) / 15

(y + 162) / 6
=
(y + 120) / 5

5(y + 162) = 6(y + 120)
5y + 810 = 6y + 720
6y − 5y = 810 − 720
y = 90
Hence, the length of the train is 90 metre.
Speed =
Distance / time

=
(90 + 162) / 18
=
252 / 18
= 14 m/sec
Converting m/sec into km/hr
14 x
18 / 5
=
252 / 5
= 50.4
Hence, the speed of the train is 50.4 km/hr

32. A train 96 metres in length passes a pole in 6 seconds and another train of the length 144 metres travelling in the same direction in 24 seconds. Find the speed of the second train.
  A.  108/5 km/hr
  B.  110/7 km/hr
  C.  118/5 km/hr
  D.  98/3 km/hr
     
   
View Answer

Shortcut:
A train l1 metres in length passes a pole in t1 seconds and another train of the length l2 metres travelling in the same direction in t2 seconds. The speed of the second train is
[
l1
t2 − t1 / t1 x t2
−
l2 / t2
]
m/sec
Here, l1 = 96, l2 = 144, t1 = 6, t2 = 24
Using these values in the shortcut, we get:
Speed of the second train =
[
96 x
24 − 6 / 6 x 24
−
144 / 24
]

=
[
96 x
18 / 144
−
144 / 24
]

= 12 − 6 = 6 m/sec
Converting into m/sec into km/hr:
6 x
18 / 5
=
108 / 5

Hence, the speed of the second train is
108 / 5
km/hr.

33. A goods train and a passenger train are running on parallel tracks in the same direction. The man of the goods train observes that the passenger train coming from behind overtakes and crosses his train in 60 seconds. Whereas a person on the passenger train finds that he crosses the goods train in 40 seconds. If the speeds of the trains be in the ratio of 1:2, find the ratio of their lengths.
  A.  1:3
  B.  3:1
  C.  1:2
  D.  2:1
     
   
View Answer

Shortcut:
A goods train and a passenger train are running on parallel tracks in the same or in the opposite direction. The driver of the goods train observes that the passenger train coming from behind overtakes and crosses his train completely in t1 seconds. Whereas a passenger on the passenger train marks that he crosses the goods train in t2 seconds. If the speeds of the trains be in the ratio of y : z, then the ratio of their lengths is given by
t2 / t1 − t2

Here, t1 = 60, t2 = 40
Using these values in the shortcut, we get:
Required ratio =
40 / 60 − 40
=
40 / 20
= 2 : 1
Hence, the ratio of their lengths is 2 : 1.

34. A goods train and a passenger train are running on parallel tracks in the opposite direction. The passenger train crosses the goods train in 40 seconds. Whereas a person in the passenger train finds that he crosses the goods train in 25 seconds. If the speeds of the trains be in the ratio of 4:5, find the ratio of their lengths.
  A.  7:3
  B.  4:5
  C.  5:3
  D.  3:5
     
   
View Answer

Shortcut:
A goods train and a passenger train are running on parallel tracks in the same or in the opposite direction. The driver of the goods train observes that the passenger train coming from behind overtakes and crosses his train completely in t1 seconds. Whereas a passenger on the passenger train marks that he crosses the goods train in t2 seconds. If the speeds of the trains be in the ratio of y : z, then the ratio of their lengths is given by
t2 / t1 − t2

Here, t1 = 40, t2 = 25
Using these values in the shortcut, we get:
Required ratio =
25 / 40 − 25
=
25 / 15
= 5 : 3
Hence, the ratio of their lengths is 5 : 3.

35. After travelling 50 km, a train meets with an accident and then proceeds at 3/4 of its former speed and reaches at its destination 35 minutes late. Had the accident occurred 24 km further, it would have reached the destination only 25 minutes late. Find the speed of the train and the distance which the train travels.
  A.  Speed 38 km/hr; Distance 144 km
  B.  Speed 48 km/hr; Distance 134 km
  C.  Speed 58 km/hr; Distance 133 km
  D.  Speed 50 km/hr; Distance 130 km
     
   
View Answer

Shortcut:
A train after travelling d1 km meets with an accident and then proceeds at
a / b
of its former speed and arrives at its destination t1 hours late. Had the accident occurred d2 km further, it would have reached the destination only t2 hours late. The speed of the train is
d2(1 − a/b) / a/b(t1 − t2)
km/hr and the distance which train travels is d1 +
d2t1 / t1 − t2
km.
Here, d1 = 50,
a / b
=
3 / 4
, t1 = 35, d2 = 24, t2 = 25
Using these values in the shortcut, we get:
Speed of the train =
24(1 − 3/4) / 3/4(35 − 25)/60

=
24(4 − 3)/4 / 3/4 x 1/6
=
6 x 4 x 6 / 3
= 2 x 4 x 6 = 48
Hence, the speed of the train is 48 km/hr.
Distance = 50 +
24 x 35 / 35 − 25

= 50 +
24 x 35 / 10
= 50 + (12 x 7) = 50 + 84 = 134
Hence, the distance that train travels is 134 km.

36. After travelling 50 km, a train meets with an accident and then proceeds at 3/4 of its former speed and reaches at its destination 25 minutes late. Had the accident occurred 24 km behind, it would have reached the destination only 35 minutes late. Find the speed of the train and the distance which the train travels.
  A.  Speed 68 km/hr; Distance 105 km
  B.  Speed 58 km/hr; Distance 100 km
  C.  Speed 40 km/hr; Distance 120 km
  D.  Speed 48 km/hr; Distance 110 km
     
   
View Answer

Shortcut:
A train after travelling d1 km meets with an accident and then proceeds at
a / b
of its former speed and arrives at its destination t1 hours late. Had the accident occurred d2 km further, it would have reached the destination only t2 hours late. The speed of the train is
d2(1 − a/b) / a/b(t1 − t2)
km/hr and the distance which train travels is d1 +
d2t1 / t1 − t2
km.
Here, d1 = 50,
a / b
=
3 / 4
, t1 = 25, d2 = 24, t2 = 35
Using these values in the shortcut, we get:
Speed of the train =
24(1 − 3/4) / 3/4(35 − 25)/60

=
24(4 − 3)/4 / 3/4 x 1/6
=
6 x 4 x 6 / 3
= 2 x 4 x 6 = 48
Hence, the speed of the train is 48 km/hr.
Distance = 50 +
24 x 25 / 35 − 25

= 50 +
24 x 25 / 10
= 50 + (12 x 5) = 50 + 60 = 110
Hence, the distance that train travels is 110 km.

37. A train running from X to Y and meets with an accident 50 km from X, after which it moves with 3/5th of of its original speed. and arrives at Y, 3 hours late. Had the accident happened 50 km further on, it would have got only 2 hours late. Find the original speed of the train and its distance from X to Y.
  A.  Speed 100/3 km/hr; Distance 200 km
  B.  Speed 100/5 km/hr; Distance 210 km
  C.  Speed 110/3 km/hr; Distance 230 km
  D.  Speed 102/3 km/hr; Distance 200 km
     
   
View Answer

Shortcut:
A train after travelling d1 km meets with an accident and then proceeds at
a / b
of its former speed and arrives at its destination t1 hours late. Had the accident occurred d2 km further, it would have reached the destination only t2 hours late. The speed of the train is
d2(1 − a/b) / a/b(t1 − t2)
km/hr and the distance which train travels is d1 +
d2t1 / t1 − t2
km.
Here, d1 = 50,
a / b
=
3 / 5
, t1 = 3, d2 = 50, t2 = 2
Using these values in the shortcut, we get:
Speed of the train =
50(1 − 3/5) / 3/5(3 − 2)

=
50(5 − 3)/5 / 3/5 x 1
=
10 x 2 x 5 / 3
=
100 / 3

Hence, the speed of the train is
100 / 3
km/hr.
Distance = 50 +
50 x 3 / 3 − 2
= 50 +
150 / 1
= 50 + 150 = 200
Hence, the distance that train travels is 200 km.

38. A train after travelling 50 km meets with an accident and then proceed at 4/5 of its former rate and arrives at the terminal 45 minutes late. Had the accident happened 20 km further on, it would have arived 12 minutes sooner. Find the rate of the train and the distance.
  A.  Speed 35 km/hr; Distance 135 km
  B.  Speed 20 km/hr; Distance 120 km
  C.  Speed 25 km/hr; Distance 125 km
  D.  Speed 30 km/hr; Distance 130 km
     
   
View Answer

Shortcut:
A train after travelling d1 km meets with an accident and then proceeds at
a / b
of its former speed and arrives at its destination t1 hours late. Had the accident occurred d2 km further, it would have reached the destination only t2 hours sooner. The speed of the train is
d2(1 − a/b) / (a/b) x t2
km/hr and the distance which train travels is
d1t2 + d2t1 / t2
km.
Here, d1 = 50,
a / b
=
4 / 5
, t1 = 45, d2 = 20, t2 = 12
Using these values in the shortcut, we get:
Speed of the train =
20(1 − 4/5) / 4/5 x 12/60

=
20(5 − 4)/5 / 4/5 x 1/5
=
4 x 1 x 5 x 5 / 4
= 5 x 5 = 25
Hence, the speed of the train is 25 km/hr.
Distance =
50 x 12 + 20 x 45 / 12
=
600 + 900 / 12
=
1500 / 12
= 125
Hence, the distance that train travels is 125 km.

39. A train meets with an accident 3 hours after starting which detains it for an hour, after which it proceeds at 75% of its original speed. It arrives at the destination 4 hours late. Had the accident taken place 150 km farther along the railway line, the train would have arrived only 7/2 hours lage. Find the length of the trip and the original speed of the train.
  A.  Speed 90 km/hr; Distance 1300 km
  B.  Speed 120 km/hr; Distance 1250 km
  C.  Speed 110 km/hr; Distance 1100 km
  D.  Speed 100 km/hr; Distance 1200 km
     
   
View Answer

Shortcut:
A train meets with an accident 't1' hours after starting, which detains it for 't' hours, after which it proceeds at
a / b
of its original speed. It arrivs at the destination 't2' hours late. Had the accident taken place 'z' km farther along the railway line, the train would have arrived only 't3' hours late. The original speed of the train is
z(1 − a/b) / a/b(t2 − t3)
km/hr and the length of the trip is
z / t2 − t3
[t2 + t1(b/a − 1) − t] km.
Here, z = 150,
a / b
=
3 / 4
, t1 = 3, t2 = 4, t3 = 7/2
Using these values in the shortcut, we get:
Speed of the train =
150(1 − 3/4) / 3/4(4 − 7/2)

=
150(4 − 3)/4 / 3/4 x(8 − 7)/2
=
75 x 1/2 / 3/4 x 1/2
=
75 x 4 / 3
= 25 x 4 = 100
Hence, the speed of the train is 100 km/hr.
Length of trip =
150 / 4 − 7/2
[4 + 3(4/3 − 1) − 1]
=
150 / (8 − 7)/2
[4 + 3(4 − 3)/3 − 1]
=
150 / 1/2
[3 + (3 x 1/3)]
=
150 / 1/2
[3 + 1]
= 150 x 2 x 4 = 1200
Hence, the distance that train travels is 1200 km.

40. A train covers a distance between stations X and Y in 45 minutes. If the speed is reduced by 5 km/hr, it will cover the same distance in 48 minutes. What is the distance between the two stations X and Y and the speed of train?
  A.  Speed 100 km/hr; Distance 80 km
  B.  Speed 80 km/hr; Distance 60 km
  C.  Speed 110 km/hr; Distance 70 km
  D.  Speed 85 km/hr; Distance 65 km
     
   
View Answer

Shortcut:
A train covers a distance between station X and Y in t1 hours. If the speed is reduced by a km/hr, it will cover the same distance in t2 hours, then the distance between X and Y is
at1t2 / (t2 − t1)
km and the speed of the train is
at2 / t2 − t1
km/hr.
Here, a = 5, t1 = 45, t2 = 48
Using these values in the shortcut, we get:
Distance between stations =
5 x 45 x 48 / (48 − 45)

=
5 x 45 x 48 / 3
= 5 x 15 x 48 = 3600
Hence, the distance is 3600 km.
Speed of train =
5 x 48 / 48 − 45

=
5 x 48 / 3
= 5 x 16 = 80
Hence, the speed of the train is 80 km/hr.

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