Question 1: A racing car covers a certain distance at a speed of 320 km/hr in 5 hours. To cover the same distance in 5/3 hours it must travel at a speed of:
Solution: Given Distance is constant.
So, Speed is inversely proportional to time.
Ratio of time 5 : 5/3 Ratio of time becomes 3 : 1 Then, Ratio of speed 1 : 3
1 unit -> 320 km/hr
3 unit -> 320 x 3 = 960 km/hr is required speed
Question 2: A train running at a speed of 36 km/hr and 100 meter long. Find the time in which it passes a man standing near the railway line is :
Solution: Speed = 36 km/hr
Change in m/s
So, speed = 36 * 5/18 = 10 m/s
Time required = Distance/speed
= 10 second
Question 3: If an employee walks 10 km at a speed of 3 km/hr, he will be late by 20 minutes. If he walks at 4 km/hr, how early from the fixed time he will reach ?
Solution: Time taken at 3 km/hr = Distance/speed
Actual time is obtained by subtracting the late time
So, Actual time = 10/3 – 1/3 = 9/3 = 3 hour
Time taken at 4 km/hr = 10/4 hr
Time difference = Actual time – time taken at 4 km/hr
= 3 – 10/4
= 1/2 hour
Hence, he will be early by 30 minutes.
Question 4: The diameter of each wheel of a truck is 140 cm, If each wheel rotates 300 times per minute then the speed of the truck (in km/hr) (take pi=22/7)
Solution: Circumference of the wheel= 2 * 22/7 * r
= 2 * 22/7 * 140/2
= 440 cm
Speed of the car = (440 * 300 * 60)/(1000 * 100 )
= 79.2 km/hr
Question 5: A man drives at the rate of 18 km/hr, but stops at red light for 6 minutes at the end of every 7 km. The time that he will take to cover a distance of 90 km is
Solution: Total Red light at the end of 90 km = 90/7 = 12 Red light + 6 km
Time taken in 12 stops= 12 x 6 = 72 minutes
Time taken by the man to cover the 90 km with 18 km/hr without stops = 90/18 = 5 hours
Total time to cover total distance = 5 hour + 1 hour 12 minute
= 6 hour 12 minute
Question 6: Two jeep start from a police station with a speed of 20 km/hr at intervals of 10 minutes.A man coming from opposite direction towards the police station meets the jeep at an interval of 8 minutes.Find the speed of the man.
Jeep Jeep + man Ratio of time 10 min : 8 min Ratio of speed 8 : 10 4 : 4+1
Here, 4 units -> 20 km/hr
1 unit -> 5 km/hr
Speed of the man = 1 unit = 1 x 5 = 5 km/hr
Question 7: Two city A and B are 27 km away. Two buses start from A and B in the same direction with speed 24 km/hr and 18 km/hr respectively. Both meet at point C beyond B. Find the distance BC.
Solution: Relative speed = 24 – 8
= 6 km/hr
Time required by faster bus to overtake the slower bus = Distance/time
Distance between B and C= 18*(27/6)= 81 km
Question 8: A man travels 800 km by train at 80 km/hr, 420 km by car at 60 km/hr and 200 km by cycle at 20 km/hr. What is the average speed of the journey?
Solution: Avg. Speed = Total distance/time taken
(800 + 420 + 200) / [(800/80) + (420/60) + (200/20)]
=>1420 / (10 + 7 + 10)
Question 9: Ram and Shyam start at the same with speed 10 km/hr and 11 km/hr respectively.
If Ram takes 48 minutes longer than Shyam in covering the journey, then find the total distance of the journey.
Speed Ratio 10 : 11 Time ratio 11 : 10
Ram takes 1 hour means 60 minutes more than Shyam.
But actual more time = 48 minute.
60 unit -> 48 min
1 unit -> 4/5
Distance travelled by them= Speed x time
= 11 x 10 = 110 unit
Actual distance travelled = 110 x 4/5
= 88 km
Question 10: A person covered a certain distance at some speed. Had he moved 4 km/hr faster, he would have taken 30 minutes less. If he had moved 3 km/hr slower, he would have taken 20 minutes more. Find the distance (in km)
Solution: Distance = [S1S2/ (S1– S2)] x T
S1 = initial speed
S2 = new speed
Distance travelled by both are same so put equal
[S (S + 4) / 4 ] * (30/60) =[ S (S – 3)/ 3 ]* (30/60)
S = 24
Put in 1st
Distance=(24 * 28) / 4 * (30/60) = 84 km
Question 11: Ram and Shyam start from the same place P at same time towards Q, which are 60km apart. Ram’s speed is 4 km/hr more than that of Shyam.Ram turns back after reaching Q and meet Shyam at 12 km distance from Q.Find the speed of Shyam.
Solution: Let the speed of the Shyam = x km/hr
Then Ram speed will be = (x + 4) km/hr
Total distance covered by Ram = 60 + 12 = 72 km
Total distance covered by Shyam = 60 – 12 = 48 km
Acc. to question, their run time are same.
72/ (x + 4) = 48/ x
72x = 48x + 192
Shyam speed is 8 km/hr
Question 12: A and B run a kilometre and A wins by 20 second. A and C run a kilometre and A wins by 250 m. When B and C run the same distance, B wins by 25 second. The time taken by A to run a kilometre is
Solution: Let the time taken by A to cover 1 km = x sec
Time taken by B and C to cover the same distance are x + 20 and x + 45 respectively
Given A travels 1000 then C covers only 750.
Distance A(1000) C(750) Ratio 4 : 3 Time 3 : 4
A/C = 3/4 = x/(x+45)
3x + 135 = 4x
Time taken by A is 2 min 15 second
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
- Time Speed Distance
- Problem on HCF and LCM
- Problem on Numbers
- Problem on Pipes and Cisterns
- Problem on Trains, Boat and streams
- Cvent Interview Experience (On campus for Internship and Full Time)
- Capgemini Interview Experience 2018 (On-Campus for Full-Time)
- American Express (On-Campus Internship, Full Time Offer)
- Add and Remove Edge in Adjacency Matrix representation of a Graph
- Check if two nodes are on same path in a tree | Set 2
- Minimum count of numbers required with unit digit X that sums up to N
- Print the longest path from root to leaf in a Binary tree
- Largest number M less than N such that XOR of M and N is even
- Shortest path in a complement graph