Problem on Time Speed and Distance

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 
= 100/10 
= 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 
= 10/3 
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. 
Solution: 

                 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 – 18 
= 6 km/hr 
Time required by faster bus to overtake the slower bus = Distance/time 
=27/6 hr 
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) 
=>1420/27 
=>1420/27 km/hr

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. 
Solution: 

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 30 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 
24x= 192 
x= 8 
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 
x =135 
Time taken by A is 2 min 15 second