BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 1

Total Questions: 45

11. In a 1000 m race Amitabh reaches the finishing line 40 seconds before Abhishek and beats him by 200 m. What is Amitabh’s speed?

Correct Answer: (b) 6.25 m/s 
Solution:

So, Abhishek takes 40 seconds to cover last 200 m i.e. his speed is 5 m/s. Now he would have taken 1000/5 = 200 s to cover the complete distance, which implies Amitabh takes 160 s, so his speed would be 1000/160 = 6.25 m/s.

12. A train overtakes two boys who are walking in the opposite direction in which the train is going at the rate of 6 km/h and 12 km/h and passes them completely in 36 seconds and 30 seconds respectively. What is the length of the train (in metres)?

Correct Answer: (b) 300 m 
Solution:

Let the length of the train be x metre, and let the speed of the train be y km/h, then
x = (y + 6) (5/18) × 36
x = (y + 12) (5/18) × 30
Solving both equations we get y = 24 km/h and x = 300 m.

13. While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5/7 of the remaining distance. What was his speed in metres per second?

Correct Answer: (a) 5/3 
Solution:Let the speed be x km/hr. Then, distance covered in 1 hr. 40 min. = 5x/3 km.
Remaining distance = (24 – (5x/3)) km.
5x/3 = 5/7 × (24 – 5x/3). Solving this, we get x = 6.
Therefore, speed = 6 km/hr = 5/3 m/s.

14. Anurag and Benoy left Chennai simultaneously towards Coimbatore. Speed of Anurag is 15 km/hr and the speed of Benoy is 12 km/hr. Half an hour later, Charles started from Chennai towards Coimbatore on the same road in the same direction. After some time he overtook Benoy and 90 minutes further, he overtook Anurag. What is the speed of Charles (in kmph)?

Correct Answer: (b) 18 
Solution:

Let Charles’s speed be x km/hr
Charles starts after half an hour
In half hour, Anurag travels = 15/2 = 7.5 km
In half hour, Benoy travels = 12/2 = 6 km
Charles catches up with Benoy after time of 6/(x – 12) hr
Charles catches up with Anurag after time of 7.5/(x – 15) hr
Given, Charles catches up with Anurag after 90 minutes of overtaking Benoy.

Hence, {7.5/(x-15)} - {6/(x-12)} = 1.5
Solving we get x = 18 km/hr

15. Ratio of speeds of Rajiv to Rahul is 4 : 5. Rahul starts moving from station A to B at 1: a.m. and Rajiv starts moving from station B to A at 2: a.m. If both meet each other at 7:00 a.m. then how much time Rajiv will take to travel from station A to station B?

Correct Answer: (d) 12.5 hours 
Solution:

Let the speed of Rajiv be 4n km/hr and speed of Rahul be 5n km/hr.
Total distance covered by Raul till 2: 00 a.m. = 5n km
Relative speed of Rahul with respect to Rajiv = (5n + 4n) = 9n km/hr
Total distance covered by both of them in 5 hours
(2 a.m. to 7 a.m.) = (9n × 5) km = 45n km
Therefore total distance between A and B = (45n + 5n) km = 50n km
So, time taken by Rajiv to travel from station A to station B = 50n/4n
= 12.5 hours

16. A can give a head start of 20 meter to B in a race of 100 meters and still both finish the race at the same time. B can give a head start of 25 meter to C in a race of 100 meters and still both finish the race at the same time. How much distance a head start of A or C can give to each other in a 100 meter race if both finishes the race at the same time?

Correct Answer: (c) A gives a head start of 40 m to C.
Solution:

According to the question; “A can give a head start of 20 meter to B in a race of 100 meters and still both finish the race at the same time.”
Hence ratio of speeds of A : B = 100 : 80 = 5 : 4…(1)
Similarly; B : C = 100 : 75 = 4 : 3…(2)
From (1) and (2) we will get; A : C = 5 : 3
It means when A can run 5 meters then in the same time C can run 3 meters.
Or when A can run 100 m then C can only run 60 m.
Hence it is clear that A can give a head start of 40 m to C.

17. David gets on the elevator at the 22nd floor of a building and rides up at the rate of 40 floors per minute. At the same time Albert gets on an elevator at the 42nd floor of the building and rides down at the rate of 60 floors per minute. If they continue travelling at these rates, then at which floor will their elevators meet?

Correct Answer: (b) 30 
Solution:Distance between 22nd floor and 42st floor = 42 - 22 = 20
Also, relative speed of both the escalators = (40 + 60) = 100 floors per minute
Therefore, time taken = 20/100 = 1/5 min or 12 sec.
Distance travelled by 1st elevator in 12 sec = 40/5 = 8 floors.
Thus, the elevators will meet at 22 + 8 = 30th floor

18. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Correct Answer: (c) 120 km/h 
Solution:Therefore, speed of train = 1.5x km/h.
Now, since the train stops for 12.5 minutes and therefore, both train and car reach the destination at same time.
So, if the train had not stopped, it would have reached 12.5 mins earlier than car, which implies it takes train to cover 12.5 mins less for the train to cover the same distance as car without stopping anywhere.
Therefore,
75/x - 75/1.5x = 12.5/60
75/x - 50/x = 5/24
25/x = 5/24
Therefore, x = 120
Speed of car = 120 km/h.

19. Sakshi was riding her cycle from home to the school. After covering 5 km she realized that the tyre had punctured due to which she had to walk the remaining distance with a reduced speed of 1/4th the original. As a result Sakshi reached her school 40 minutes late. Had the tyre gotten punctured after riding for 8 km, then she would have reached 10 minutes late. What is the distance between Sakshi’s home and her school?

Correct Answer: (a) 9 km 
Solution:Let the initial speed and distance be s and d, so ATQ
So d/s + 40/60 = (d-5)/(1/4)s + 5/s …(1)
d/s + 10/60 = (d-8)/(1/4)s + 8/s …(2)
Solving, 1 and 2 together, we get
D = 9 and s = 18

20. Akshay and Bobby are running on a circular track of radius 147 metres. Bobby can complete a round in 84 seconds and the speed of Akshay is half the speed of Bobby. They started simultaneously towards each other from two points 294 metres diametrically opposite on the circular path. If they first meet at a point which is between the two points from where they started their race, after how much time from the start do they meet at that point for the fifth time?

Correct Answer: (c) 700 seconds 
Solution:

Circumference of the track = 2 × (22/7) × 147 = 924 m.
Distance to be covered for the first meeting (by both of them) = 462 m.

Speed of Akshay = 924/168 = 5.5 m/s.
Speed of Bobby = 11 m/s.
Time taken from the start to the first meeting
= 462/(16.5) = 924/33 = 28s.
Time taken by Akshay and Bobby to meet again at that point 28 + 84×8 = 700 seconds.