# 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 at speed of 10 km at 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 pie=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: **

Bus Bus + manRatio 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 18 km/hr and 24 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

=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, 400 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 + 400 + 200) / [(800/80) + (420/60) + (200/20)]

=>1400 / (10 + 70 + 10)

=>1400/90

=>**140/9 km/hr**

**Question 9: ** Ram and Shyam start at the same with speed 10 km/hr and 11 km/hr respectively.

If A takes 48 minutes longer than B 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 20 minutes more. Find the distance (in km)

**Solution: **** Distance = [S _{1}S_{2}/ (S_{1}– S_{2})] x T**

S

_{1}= initial speed

S

_{2}= 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. 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**

## Recommended Posts:

- 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)
- American Express (On-Campus Internship, Full Time Offer)
- Capgemini Interview Experience 2018 (On-Campus for Full-Time)
- What Should be your Best Strategy for this Placement Season?
- Placement 100 : Complete Interview Preparation package you must have this Placement Season
- Increase your salary as Software Developer : Learn Programming from Industry Experts here
- HCL Placement Paper | Quantitative Aptitude Set - 5
- HCL Placement Paper | Quantitative Aptitude Set - 3
- IBM Placement Paper | Verbal Reasoning Set - 2