Direction (1-5): Study the following data carefully and answer the questions:

Pie chart given below shows the distribution (degree) of time taken (seconds) by five trains to cross a pole. Sum of the pie chart is 150 seconds

Table given below shows the length of train more than the length of a platform.

Note: Time taken by train A to cross the platform is 54 seconds.

1) What is the time taken by train A to cross a train P of length 150 meters running in same direction as that of train A with speed 7 km/h?

a) 1.5 minute

b) 2 minute

c) 45 seconds

d) 1 minute

e) 75 seconds

2) If train B crosses a platform of certain length in 30.6 seconds, then what is the length of the platform?

a) 175 m

b) 225 m

c) 250 m

d) 275 m

e) 300 m

3) Train C crosses a person running in opposite direction in 20.25 seconds, then what is the time taken by train B to cross the same person running in same direction?

a) 24 s

b) 20 s

c) 18 s

d) 16 s

e) 28 s

4) What is the time taken by train E to cross D when running in same direction?

a) 132.8 s

b) 144.4 s

c) 175.2 s

d) 116.8 s

e) 192.4 s

5) Train M length is average lengths of all the five trains. Train M crosses a train E opposite direction in 24 seconds, then what is the speed of train M?

a) 46.8 s

b) 72.8 s

c) 62.4 s

d) 59.4 s

e) 86.4 s

Answers:

Direction (1-5)

Time taken by train A to cross Pole = 150 * (86.4/360) = 36 seconds

Time taken by train B to cross Pole = 150 * (43.2/360) = 18 seconds

Time taken by train C to cross Pole = 150 * (51.84/360) = 21.6 seconds

Time taken by train D to cross Pole = 150 * (115.2/360) = 48 seconds

Time taken by train E to cross Pole = 150 * (63.36/360) = 26.4 seconds

Let length of platform = x

Length of train A is ‘x + 200’.

Let speed of train A is ‘A’.

Now,

(x + 200) = A * 36 * (5/18)

(x + 200) = 10A ………. (1)

(x + 200 + x) = A * 54 * (5/18)

(2x + 200) = 15A ……. (2)

From (1) and (2):

(2x + 200) = 1.5(x + 200)

2x + 200 = 1.5x + 300

0.5x = 100

x = 200 and A = 40

Length of platform = x = 200 meters

Now,

Train B

(x + 50) = B * 18 * (5/18)

250 = 5B

B = 50 km/h

Train C

(x + 250) = C * 21.6 * (5/18)

450 = 6C

C = 75 km/h

Train D

(x + 200) = D * 48 * (5/18)

400 = 40D/3

D = 30 km/h

Train E

(x + 130) = E * 26.4 * (5/18)

330 = 22E/3

E = 45 km/h

Table given below shows the length (meters) and speed (km/h) of trains:

1) Answer: D

Let required time = ‘t’ seconds

(400 + 150) = (40 – 7) * t * (5/18)

550/33 = 5t/18

t = 60 seconds

t = 1 minute

2) Answer: A

Let the length of platform = ‘x’ meters

Now,

(250 + x) = 50 * 30.6 * (5/18)

250 + x = 425

x = 175 meters

3) Answer: B

Let the speed of person = ‘x’ km/h

Now,

450 = (75 + x) * 20.25 * (5/18)

450 = 5.625 * (75 + x)

80 = 75 + x

x = 5 km/h

Let time taken by train B to cross the person = ‘t’ seconds

250 = (50 – 5) * t * (5/18)

250 = 12.5t

t = 20 seconds

4) Answer: C

Let required time = ‘t’ seconds

Now,

(400 + 330) = (45 – 30) * t * (5/18)

730 = 25t/6

t = 175.2 seconds

5) Answer: D

Length of train M = (400 + 250 + 450 + 400 + 330)/5 = 366 meters

Let the length of train M = ‘x’ meters

Now,

(330 + 366) = (45 + x) * 24 * (5/18)

696 = 20(45 + x)/3

104.4 = 45 + x

x = 59.4 seconds