Deloitte Problems on Trains Questions and Answers Quiz-1

Solution:

Speed = 82 \times \frac{5}{18} m/sec = 22.7 m/sec.

Length of the train = Speed \times Time.

Length of the train = 22.7 x 14 m = 317.8 m = 318 m

Solution:

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

Then, length of the first train = 32x metres,

Length of the second train = 25y metres.

\frac{32x + 25y}{x +y} = 28

32x + 25y = 28x +28y

4x = 3y

\frac{x}{y} = \frac{3}{4}

 

Solution:

Relative speed = (200 + 100) km/hr = 300 km/hr

300 \times \frac {5}{18} m/sec = 83.33 m/sec.

Let the length of the other train be x metres.

Then,\frac{x + 350}{15} = 83.33

x + 350 = 1249.95

x = 949.95 m = 950 m (approx)

Solution:

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.

\frac{550 + 550}{8} = 3x

24x = 1100

x = 45.8

So, speed of the faster train = 45.8 m/sec

= 45.8 \times \frac{18}{5} km/hr

= 164.88 km/hr = 165 km/hr (approx)

Solution:

Speed = \frac{450}{21}m/sec = \frac{150}{7} m/sec

Let the length of the platform be x metres.

Then, \frac{x + 450}{50} = \frac{150}{7}

7(x + 450) = 7500

x = 621.42 m = 621 m (approx)

Solution:

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

Then, \frac{x}{y} = 10

x = 10y

Now, \frac{x + 342}{20} = y

=> 10y + 264 = 20y

=> y = 26.4

Therefore Speed = 26.4 m/sec = 26.4 \times \frac{18}{5}  km/hr = 95 km/hr.

Solution

Speed = \frac{Distance}{Time}

Relative speed = Speed of train + Speed of boy

60 = Speed of train + 25

Speed of train = 60 – 25 = 35 km/hr

35km/hr = 35 \times \frac{5}{18} = 9.72 m/s

Distance = Speed x Time = 9.72 x 20 = 194.4 m

Solution:

Let the speed of faster train = X m/sec

Let the speed of slower train = Y m/sec

Then, relative speed when they are moving in opposite direction = X + Y

X + Y = \frac{145 + 175}{20}

X + Y = 16.....(i)

Relative speed when they are moving in the same direction

X - Y = \frac{145 + 175}{80}

X - Y = 4....(ii)

On solving (i) and (ii) simultaneously, we get x = 10 m/sec

substituting the value of x in any one  of the equation, we get y = 6m/s

Solution:

Speed of train relative to jogger = (65 - 15) km/hr = 50 km/hr.

= 50 \times \frac{5}{18}  = \frac{125}{9} m/sec.

Distance to be covered = (280 + 180) m = 460 m.

Time taken = 460\times \frac {9}{125} sec = 33.12 sec.

Solution:

Speed = 54 \times \frac {5}{18} = 15 m/sec.
Time = 30 sec.

Let the length of the train be x metres.

Then, \frac{x + 240}{30} = 15

x + 250 = 450

x = 200 m.