BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 2

Total Questions: 45

21. The average speed of a train in the onward journey is 10% more than that in the return journey. The total time taken for the complete to and fro journey is 16 hours, covering a distance of 2000 km. The speed of the train in the onward journey?

Correct Answer: (b) 131 km/hr 
Solution:From the given data, we get
Let the speed in return journey be x km/hr
⇒ speed in onward journey = 110/100 × x = 11x/10 km/hr
⇒ Average speed = (2 × S1 × S2)/(S1 + S2) = (2 × 11x/10 × x)/(11x/10 + x) = 22x/21 km/hr
⇒ (2000 × 21)/(22x) = 16
⇒ x = 119 km/hr
∴ Speed in onward journey = (11/10 × x) = (11/10 × 119) = 130.9 km/hr ≅ 131 km/hr (approximately)

22. Find the length of the train if it takes 20 seconds to pass through platform A which is 200 m long and 15 seconds to pass through another platform B which is 100 m long

Correct Answer: (d) 200 
Solution:

Let the length of the train be x m
⇒ Distance travelled by train when it pass platform A = (200 + x) m
⇒ Distance travelled by train when it pass platform B = (100 + x) m
⇒ Speed of the train when it pass platform A = (200 + x)/20
⇒ Speed of the train when it pass platform B = (100 + x)/15
∴ Speed of train remain same
∴ (200 + x)/20 = (100 + x)/15
⇒ 15(200 + x) = 20(100 + x)
⇒ 3000 + 15x = 2000 + 20x
⇒ x = 1000/5 = 200
∴ The length of the train is 200 m

23. Two trains of 132 m and 108 m in length run on parallel lines of tracks. When travelling in opposite direction they pass each other in 5 seconds but when they are running in the same direction with the same speed as before, the faster train passes the other in 10 seconds. Find their approximate speeds. (in Km/Hr)

Correct Answer: (d) 130, 43 
Solution:When trains are moving in opposite direction;
relative speed = 132 + 108/5
⇒ 48 m/sec = S1 + S2
When trains are moving in the same direction;
relative speed = 132 + 108/10
⇒ 24 m/sec = S1 - S2
∴ Speed of faster train;
S1 = (48 + 24)/2 = 36 m/sec = 36 × 18/5 km/hr
⇒ 129.6 km/hr ≅ 130 km/hr
And speed of slower train;
S2 = (48 - 24)/2 = 12 m/sec = 12 × 18/5 km/hr
⇒ 43.2 km/hr ≅ 43 km/hr
∴ Their speeds are 130 km/hr and 43 km/hr

24. A train leaves a station A at 5 p.m. and reaches another station B at 9 p.m., another train leaves B at 6 p.m. and reaches A at 9 : 30 p.m. Find the time when the two train crosses each other?

Correct Answer: (d) 07 : 24 p.m.
Solution:

Let the distance between station A & station B = 140 km (Just assume any number which will make our Calculation easy)
A train leaves a station A at 5 p.m. and reaches another station B at 9 p.m.
Time taken by first train = 9 - 5 = 4 hours
Another train leaves B at 6 p.m. and reaches A at 9 : 30 p.m.
Time taken by second train = 9.5 - 6 = 3.5 hours
Speed of first train = 140/4 = 35 km/h

Speed of second train = 140/3.5 = 40 km/h
Till 6:00 p.m. first train would have covered 35 km
and second train will start from now.
Remaining distance = 140 - 35 = 105
Sum of speed = 40 + 35 = 75 km/h
Time = 105/75 = 1.4 hours = 1 hour and 24 minutes
Meeting time of the Train 06:00 p.m. + (1 hour and 24 minutes) = 07:24 p.m.

25. Suresh walks from his home to Bus station. If he walks at the speed of 9 kmph, he reaches the bus station 3 minutes before the arrival of the bus .However, if he walks at the speed of 6 kmph, he misses the bus by 2 minutes. Find the distance covered by him to reach the bus station.

Correct Answer: (a) 1.5 km 
Solution:

1/6 + 1/9) = 3/18 = 1/6
(3d - 2d)/18 = 1/12
D = 18/12 = 1.5 km

26. Anand usually travels from city ‘A’ to ‘B’ at 24 km/h. On a certain day he travelled 40% of the distance at 125% of his usual speed and remaining distance at 75% of his usual speed such that he took 1.5 hours more than usual. Find the distance between city ‘A’ and ‘B’.

Correct Answer: (a) 300 km 
Solution:Let the distance between city ‘A’ and ‘B’ be = ‘10d’ km
According to the question,
{(10d × 0.40) ÷ (1.25 × 24)} + {(10d × 0.60) ÷ (0.75 × 24)} - (10d ÷ 24) = 1.5
Or, (4d/30) + (6d/18) - (10d/24) = 1.5
Or, (0.8d + 2d - 2.5d) ÷ 6 = 1.5
Or, 0.3d = 9
So, d = 9 ÷ 0.3 = 30
So, total distance travelled by Anand = 10 × 30 = 300 km
Hence, option a.

27. A car running with a speed of 60 km/hr takes 8 hours to go from ‘A’ to ‘B’. If on the way back from ‘B’ to ‘A’, the car takes 25% less time than it takes while going from ‘A’ to ‘B’, then find the approximate average speed of the car during the whole journey.

Correct Answer: (a) 68.6 km/h 
Solution:Total distance covered by the car in 8 hours = 60 × 8 = 480 km
Time taken by the car in the return journey = 0.75 × 8 = 6 hours
Average speed = Total distance covered ÷ Total time taken
Average speed of the car = (480 × 2) ÷ (8 + 6) ≈ 68.6 km/h
Hence, option a.

28. Poorva express and Maurya express started from Delhi and Howrah, respectively towards each other at the same time. After 16 hours, when they meet each other, Maurya express has travelled 256 km more than Poorva express. If the distance between Delhi and Howrah is 1280 km, then find the distance travelled by Poorva express with same speed in 10 hours.

Correct Answer: (a) 320 km 
Solution:Let distance travelled by Poorva express = ‘x’ km
So, distance travelled by Maurya express = ‘x + 256’ km
ATQ,
x + x + 256 = 1280
2x + 256 = 1280
2x = 1024
So, x = 512
Distance travelled by Poorva express = 512 km
So, speed of Poorva express = 512/16 = 32 km/hr
Therefore, distance travelled by Poorva express in 10 hours = 32 × 10 = 320 km
Hence, option a.

29. A train takes 17.6 seconds to cross a pole and 28 seconds to cross a 260-metre-long platform. A car crosses a 315-metre-long bridge in 15 seconds. Find the time taken by the train to completely overtake the car.

Correct Answer: (e) 110 seconds
Solution:

): Let the length of the train = ‘d’ metres
Then, distance travelled by the train in 17.6 seconds = ‘d’ metres
Distance travelled by the train in 28 seconds = (d + 260) metres
So, distance travelled by the train in (28 - 17.6) seconds = d + 260 - d = 260 metres
So, speed of the train = 260 ÷ 10.4 = 25 m/s
And so, length of the train = 25 × 17.6 = 440 metres
Speed of the car = 315 ÷ 15 = 21 m/s

So, relative speed of the train w.r.t the car
= 25 - 21 = 4 m/s
And so, time taken by the train to completely overtake the car = 440 ÷ 4 = 110 seconds
Hence, option e.

30. In a 240-metre race, ‘A’ beats ‘B’ by 48 metres, and ‘B’ beats ‘C’ by 15 metres. By how much distance does ‘A’ will beat ‘C’, in the same race?

Correct Answer: (b) 60 metres 
Solution:Ratio of speeds/distance of ‘A’ and ‘B’ = 240:192
= 5:4
Ratio of speeds/distance of ‘B’ and ‘C’ = 240:225
= 16:15
So, ratio of speeds of ‘A’, ‘B’ and ‘C’ = 20:16:15
So, distance covered by ‘C’ by the time ‘A’ covers 240 metres = 240 ÷ 20 × 15 = 180 metres
So, ‘A’ beats ‘C’ by 240 - 180 = 60 metres
Hence, option b.