1705.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
Explanation:
| Speed =$ \left(\dfrac{240}{24} \right) $m/sec = 10 m/sec. |
| $\therefore$ Required time =$ \left(\dfrac{240 + 650}{10} \right) $sec = 89 sec. |
1706.How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
25
30
40
45
Explanation:
| Speed of the train relative to man | = (63 - 3) km/hr |
| = 60 km/hr | |
| =$ \left(60 \times\dfrac{5}{18} \right) $m/sec | |
| =$ \left(\dfrac{50}{3} \right) $m/sec. | |
| $\therefore$ Time taken to pass the man | =$ \left(500 \times\dfrac{3}{50} \right) $sec |
| = 30 sec. |
1707.A train has a length of 150 meters it is passing a man who is moving at 2 km/hr in the same direction of the train in 3 seconds. Find out the speed of the train.
182 km/hr
180 km/hr
152 km/hr
169 km/hr
Explanation:
Length of the train, l = 150m
Speed of the man , Vm= 2 km/hr
Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr
Relative Speed = Speed of train, Vt - Speed of man
=> 180 = Vt - 2
=> Vt = 180 + 2 = 182 km/hr
1708.A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
5 sec
6 sec
7 sec
10 sec
Explanation:
Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.
| =$ \left(66 \times\dfrac{5}{18} \right) $m/sec |
| =$ \left(\dfrac{55}{3} \right) $m/sec. |
| $\therefore$ Time taken to pass the man =$ \left(110 \times\dfrac{3}{55} \right) $sec = 6 sec. |
1710.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other is:
9
9.6
10
10.8
Explanation:
| Relative speed = (60 + 40) km/hr =$ \left(100 \times\dfrac{5}{18} \right) $m/sec=$ \left(\dfrac{250}{9} \right) $m/sec. |
Distance covered in crossing each other = (140 + 160) m = 300 m.
| Required time =$ \left(300 \times\dfrac{9}{250} \right) $sec=$ \dfrac{54}{5} $sec = 10.8 sec. |
1713.A train travelling at a speed of 75 mph enters a tunnel 3 $\dfrac{1}{2}$ miles long. The train is $\dfrac{1}{4}$ 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?
2.5 min
3 min
3.2 min
3.5 min
Explanation:
| Total distance covered | =$ \left(\dfrac{7}{2} +\dfrac{1}{4} \right) $miles |
| =$ \dfrac{15}{4} $miles. |
| $\therefore$ Time taken | $ =\left(\dfrac{15}{4\times 75}\right)hrs$ |
| $=\dfrac{1}{20}hrs $ | |
| $ =\left(\dfrac{1}{20}\times 60\right)min.$ | |
| = 3 min. |
1717.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
18 sec
36 sec
72 sec
Explanation:
Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.
| =$ \left(36 \times\dfrac{5}{18} \right) $m/sec |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
1720.Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres in what time (in seconds) will they cross each other travelling in opposite direction?
10
12
15
20
Explanation:
| Speed of the first train =$ \left(\dfrac{120}{10} \right) $m/sec = 12 m/sec. |
| Speed of the second train =$ \left(\dfrac{120}{15} \right) $m/sec = 8 m/sec. |
Relative speed = (12 + 8) = 20 m/sec.
| $\therefore $Required time =[$ \dfrac{(120 + 120)}{20}] $sec = 12 sec. |
1721.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:
50 m
72 m
80 m
82 m
Explanation:
Let the length of each train be $ x $ metres.
Then, distance covered = 2$ x $ metres.
Relative speed = (46 - 36) km/hr
| =$ \left(10 \times\dfrac{5}{18} \right) $m/sec |
| =$ \left(\dfrac{25}{9} \right) $m/sec |
| $\therefore \dfrac{2x}{36} $=$ \dfrac{25}{9} $ |
$\Rightarrow$ 2$ x $ = 100
$\Rightarrow x $ = 50.
1723.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
23$ \dfrac{2}{9} $m
27$ \dfrac{7}{9} $m
29 m
Explanation:
| Relative speed = (40 - 20) km/hr =$ \left(20 \times\dfrac{5}{18} \right) $m/sec =$ \left(\dfrac{50}{9} \right) $m/sec. |
| $\therefore$ Length of faster train =$ \left(\dfrac{50}{9} \times 5\right) $m =$ \dfrac{250}{9} $m = 27$ \dfrac{7}{9} $m. |