BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 1

Total Questions: 45

21. A criminal sees a jeep at a distance of 400m coming towards him at 54 km/hr. The criminal takes 10 seconds to realize that it’s a police jeep and starts his bike, running away from the police at a speed of 72 km/hr. The police take 20 seconds from the start to realize that a criminal is running away and increase the jeep’s speed to 90 km/h. How long after the criminal saw the police jeep did police catch him?

Correct Answer: (d) 70 seconds
Solution:

Initial distance of police from the criminal = 400m.
In 10 seconds, police covers 150 meters.
Hence the distance of police from the criminal after 10 seconds = 250m.
In the next 10 seconds, relative speed of the criminal = (20-15) = 5 m/s.
The distance become 300 metres.
After a total of 20 seconds, relative speed of police to the criminal = (25-20) = 5 m/s.
Hence time taken to catch him = 60 seconds.
Total time elapsed = 80 seconds.
Therefore, police jeep caught the criminal after 70 seconds of the criminal seeing the jeep.

22. A car is climbing up a hill at a speed of 90 km/h. A rabbit, sitting at the top of the hill, spots the car when it is 1170 km away, and start running down the hill towards car at a speed of 140 km/h. As soon as it meets the car, the rabbit turns back towards the top of the hill at a speed of 120 km/h. The rabbit continues this to and fro motion from the top of the hill towards the car and again back at the top of the hill till the car reaches the top of the hill. Find the total distance covered by the rabbit?

Correct Answer: (e) 1680 km
Solution:Total time taken by car to reach the top of the hill = 1170/90 = 13 hours.
Average speed of the rabbit = (2 × 140 × 120)/(140 + 120) = 1680/13 km/h
Total distance covered by the rabbit = (1680/13 × 13) = 1680 km

23. Shobhit decides to climb a mountain. On the first day, he climbs 1/10th of the mountain. On the second day, he climbs 2/3rd of the climb he made on the first day. Next day, he covers 1/10 of the remaining distance, on the fourth day, he covers 2/3rd of the distance he already climbed. He continues climbing with the same pattern of day 3rd for fifth day and of sixth day as of fourth day, for a total of six days. At the end of the sixth day, he finds that 5 km more would see the end of his journey, find the total distance that needs to be covered to reach the top of the mountain.

Correct Answer: (e) 24
Solution:

First day, he covers 1/10 of the distance, second day he covers 1/15, third day he covers 1/12, fourth day he covers 1/6 of the distance, fifth day he covers 7/120 of the distance and on sixth day he covers 19/60 of the distance.
Adding all the distances, he covers 95/120 of the total distance in 6 days.
Remaining distance = 25/120 = 5 kilometres.
Thus total distance = 24 km.

24. The Delhi-Howrah Duranto Express left Delhi for Howrah. Having travelled 1200 km without stopping in between, which constitutes 80% of the distance between Delhi and Howrah, the train was stopped at a technical halt. When the train started again an hour later, its speed was increased by 15 km per hour to make it reach the destination on time. Find the initial speed of the Duranto Express.

Correct Answer: (a) 60 kmph 
Solution:80% of the distance = 1200 km
Total distance = 1500 km
Let the initial speed of the train be x km per hour.
Scheduled time = 1500/x hours
Last 300kms are travelled at x+15 km per hour.
Thus 1500/x = 1200/x + 300/(x+15) + 1
=> x = 60 km per hour.

25. A club owner drives his car from his club to his airport at an average speed of 30 mph (miles per hour). At the airport a chopper was waiting for him. He flew from the airport to Dubai at an average speed of 40mph. It took 4 hours for the entire trip. The entire distance he covered was 150 miles. Find the distance from the airport to the club.

Correct Answer: (e) 30 miles
Solution:

Let the distance from the airport to club be ‘x’.
The distance he travel in chopper will be ‘150 - x’
T1 = D1/S1 = (150 - x)/40
T2 = D2/S2 = x/30
T1 + T2 = T3
(150 - x)/40 + x/30 = 4
450 - 3x + 4x = 480
x = 30 miles

26. Bob, Damon and Charlie decided to race on a circular race track. All of them start from the same point at different times. Damon started 3 minutes after Charlie and both Bob and Damon goes ahead of Charlie's car at 8 pm on the same day. The speed of Bob, Damon and Charlie's cars are 120, 80, 70 metre per minute respectively. Find the time at which Bob started?

Correct Answer: (a) 7:46 pm 
Solution:

Distance covered by Charlie in the 1st 3 minutes = 210 m.
Time taken by Damon to overtake Charlie:
Let the distance travelled by them until they meet = x + 210
x/70 = (x+210)/80
x = 1470
Total distance travelled = 1470 + 210 = 1680

Time in which Bob overtook Charlie = 1680/120
= 14 minutes
So, time at which Bob started = 7:46 pm

27. Adam starts walking at 9:00 hrs towards a certain destination point at a speed of 6kmph. Ben starts cycling towards the same destination point at 11:30 hrs at a speed of 8kmph. If they continue to walk and cycle at the same pace, at what time will the distance covered by both of them be same?

Correct Answer: (b) 19:00 hrs 
Solution:Considering option (b):

Adam starts walking at 9:00, so every one hour he completes 6km.
By 19 hrs in the evening he completes 60 km.
Now, Ben starts cycling at 11:30 and completes 8km in one hour.
By 18:30 in the evening he completes 56kms.
In another 30 min he travels 4 km.
Thus by 19 hrs in the evening he also travels 60 km

28. Raju, a labourer decides to go to a week - long adult literacy camp set up by the government. At what time will Raju reach the camp on the fourth day if he starts cycling at 10:15 A.M from his home and given the fact that the camp keeps shifting 2 Km farther away from Raju's house each day. Raju's cycling speed is 12Km/hr and the time taken by Raju to reach the camp on the 2nd day was 3hrs.

Correct Answer: (a) 1:35 P.M. 
Solution:

Let the distance between Raju’s house and the camp on the 1st day be x Km.
=> The distance between Raju’s house and the camp on the 2nd day be (x+2) Km.
Time taken by Raju on the 2nd day = 3hrs.
(x+2) / 3 = 12
=> X + 2 = 12 × 3 => X + 2 = 36
=> X = 34Km

Time taken by Raju to reach the camp on the fourth day = (x + (3×2)) / 12
=> [34 + 6] / 12 => 40/12
=> 10/3 hrs
Now 1 hr = 60 minutes
=> (10/3) hrs = (10/3) × 60 minutes
=> 200 minutes
=> 3 hrs and 20 minutes
=> He will reach camp at 3 hrs and 20 minutes from 10:15 A.M
=> 1:35 P.M

29. Abhi and Bunny travels from P to Q via R. Abhi travels from P to R in 4 hours and from R to Q in 5 hours. Bunny travels from P to R then R to P and then P to R in 7 hours and from R to Q in 5 hours. If the average of Abhi and Bunny for the whole journey is same, find the ratio of the distance between P to R to the distance between R to Q.

Correct Answer: (d) 1/5 
Solution:

Let the distance from P to R be x and from R to Q be y
Total distance travelled by Abhi = x + y
Total time taken by Abhi = 4 + 5 = 9hrs
Total distance travelled by Bunny = 3x + y
Total time taken by Bunny = 7 + 5 = 12hrs
Average speed of Abhi = (x+y)/9
Average speed of Abhi = (3x+y)/12
Average speed of Abhi and Bunny is equal
(x+y)/9 = (3x+y)/12
=> (x+y)/3 = (3x+y)/4
=> 4x+4y = 9x+3y
=> y = 5x
Ratio of the distance between P to R to the distance between R to Q = x/y = 1/5

30. Mohit travels daily to his office situated 100 km away. On one particular day, he travels the first half of his way at his usual speed. After that, he had to stop for exactly 10/3 minutes due to a minor accident on the road. To make up for the lost time, he increased his speed by 10 kmph. Find the usual speed (in km per hour) with which Mohit travels to his office.

Correct Answer: (e) 90
Solution:Let the usual speed of Mohit be v km per hour.
He travels the first 50 km with this speed.
He travels the next 50 km with speed v+10, and takes 10/3 minutes less than if he had travelled with his usual speed.
Thus, (Time taken at v) – (time taken at v+10) = 1/18 hours.
50/v – 50/(v+10) = 1/18
50×10/(v×(v+10)) = 1/18
=> v² + 10v - 9000 = 0
=> v = 90, -100
Therefore, his usual speed = 90 km per hour.