Question 1 |

Samuel covers the distance from his home to his office at a speed of 25 km/hr and comes back at a speed of 4 km/hr. He completes the whole journey within 5 hours 48 minutes. Find out the distance from his home to office:

20 km | |

18 km | |

15 km | |

25 km |

**Time Speed Distance**

**Discuss it**

Question 1 Explanation:

Let the speed of travelling to office and back to home be x and y respectively.
So, his average speed is = 2xy / (x+y) = (2 × 25 × 4) / (25 + 4) = 200/29 km/hr
He covers the whole journey in 5 hours 48 minutes = 5⅘ = 29/5 hrs
Therefore, total distance covered = (200/29 × 29/5) = 40 km
So, the distance from his home to office = 40/2 = 20 km

Question 2 |

If John walks at the speed of 5 km/h, he reaches his office 7 minutes late. However, if he walks at the speed of 6 km/h, he reaches his office 5 minutes early. How far is his office from his home?

9 | |

8 | |

10 | |

6 |

**Time Speed Distance**

**Discuss it**

Question 2 Explanation:

Let the distance of John’s office from his home be x.
The time difference when covering the distance x at the two different speeds = 5 - (-7) = 12 min = 1/5 hr
⇒ x/5 - x/6 = 1/5
⇒ (6x - 5x)/30 = 1/5
⇒ x = 6.
So, his office is 6 km far from his home.

Question 3 |

A policeman sees a thief at a distance of 100 meters and starts to chase him. The thief sees him and starts to run too. If the thief is running at the speed of 8 km/hr and the policeman is running at the speed of 10 km/hr, find out the distance covered by the thief before the policeman catches him.

250 meters | |

400 meters | |

450 meters | |

401 meters |

**Time Speed Distance**

**Discuss it**

Question 3 Explanation:

We can safely assume that the policeman is running in the same direction as the thief.
Speed of policeman w.r.t thief = (10 - 8) = 2 km/hr.
Time taken by policeman to cover the 100m distance between him and the thief = (100/1000) / 2 = 1/20 hr.
Therefore, the distance covered by thief in 1/20 hrs = 8 × 1/20 = 2/5 km = 400 meters.

Question 4 |

Paul has to travel 24 km. After walking for 1 hour 40 minutes he sees that he has covered 5/7 of the distance left to cover. Find out Paul’s speed in meters per second.

5/3 m/s | |

7/5 m/s | |

2/3 m/s | |

8/5 m/s |

**Time Speed Distance**

**Discuss it**

Question 4 Explanation:

Let the required speed be x km/hr.
Distance covered by Paul in 1 hr 40 min = x × 100/60 = 5x/3 km.
Remaining distance = (24 - 5x/3) km.
Therefore, 5x/3 = 5/7 × (24 - 5x/3)
⇒ 7/5 × 5x/3 = 24 - 5x/3
⇒ 7x/3 = (72 - 5x)/3
⇒ 7x = 72 - 5x
⇒ 12x = 72 ⇒ x = 6.
Paul's speed in meters per second = 6 × 5/18 = 5/3 m/s.

Question 5 |

The ratio of the speed of two trains is 7:8. If the second train covers 400 km in 4 h, find out the speed of the first train.

69.4 km/h | |

78.6 km/h | |

87.5 km/h | |

40.5 km/h |

**Ratio and Proportion**

**Time Speed Distance**

**Discuss it**

Question 5 Explanation:

Let the speed of the two trains be 7x and 8x.
Then, 8x = 400 / 4
⇒ 8x = 100 ⇒ x = 12.5 km/h.
Hence, speed of the first train = 7x = 7 × 12.5 = 87.5 km/h.

Question 6 |

Rajdhani Express halts for 3 minutes every time it covers a distance of 75 km. If the train runs at a speed of 100 km/h and the destination is 600 km away from the source, find out the time taken to reach the destination station from the source station.

6 h 23 min | |

6 h 22 min | |

6 h 21 min | |

6 h 24 min |

**Time Speed Distance**

**Discuss it**

Question 6 Explanation:

Since the train runs at a speed of 100 km/h, the time taken to cover 600 km = 600/100 = 6 h.
Number of times the train halts = 600/75 - 1 = 7.
Since the train halts for 3 minutes at each stop, the time spent waiting = 7 * 3 = 21 min.
Therefore, total time taken = 6 h 21 min.

Question 7 |

Max completes his journey at an average speed of 9 km/h. He covers the first 9 km at a speed of 6 km/h and he takes 1·5 hours to cover the remaining distance. Find out the speed at which he covered the remaining distance.

11 km/h | |

12 km/h | |

13 km/h | |

15 km/h |

**Time Speed Distance**

**Discuss it**

Question 7 Explanation:

Let the required speed be x km/h.
Total time taken to finish his journey = (9/6 + 1·5) = 3 hours.
Total distance = 9 + 1·5x km.
Given, average speed = 9 km/h.
Therefore, (9 + 1·5x)/3 = 9
⇒ 9 + 1·5x = 27
⇒ 1·5x = 18
⇒ x = 12 km/h.

Question 8 |

Peter and Beckon start to walk in the same direction together. If Peter's speed is 5 km/h and Beckon's speed is 6 km/h, find out the time duration after which they are 17 km apart.

17 | |

15 | |

19 | |

20 |

**Time Speed Distance**

**Discuss it**

Question 8 Explanation:

In 1 hour Peter covers 5 km and Beckon covers 6 km.
So, they are 1 km apart after 1 hour.
Therefore, they are 17 km apart after 17 hours.

Question 9 |

A car covers a distance of 450m in 90 secs. What is the speed in km/hr?

20 | |

5 | |

10 | |

18 |

**Time Speed Distance**

**Discuss it**

Question 9 Explanation:

Speed = distance/time
= 450m/90sec = 5m/sec
= (5*60*60)/(1000) = 18 Km/hr.

Question 10 |

A train covers a journey of 4 stations connected to form a square at speeds of 20, 40, 60 and 80 km/hr. What is the average speed of train for this journey?

38.4 | |

100 | |

80.4 | |

160 |

**Time Speed Distance**

**Discuss it**

Question 10 Explanation:

Speed = distance/time
Say, side of square = y km
time t1 = y/20
t2 = y/40, t3 = y/60 and t4 = y/80
Total time = (12+6+4+3)y/240
speed = total distance/total time
= 4y/(t1+t2+t3+t4) = 4y*240/25y = 38.4 km/hr

Question 11 |

A car completes a journey in 10 hrs, the first half at 11 kmph and the second half at 14kmph. Find the total distance covered?

120.5 | |

123.2 | |

130.4 | |

134.5 |

**Time Speed Distance**

**Discuss it**

Question 11 Explanation:

Average speed = 2V1*V2/(V1+V2)

= 2*11*14/25

Distance = (22*14/25)*10

= 44*14/5

= 123.2 km.

= 2*11*14/25

Distance = (22*14/25)*10

= 44*14/5

= 123.2 km.

Question 12 |

The speed of a child while going to school is 2 kmph and returns to the village at a speed of 3 kmph. If he takes 5 hrs in all, what is the distance between the village and the school?

4 Km | |

6 Km | |

8 Km | |

10 Km |

**Time Speed Distance**

**Discuss it**

Question 12 Explanation:

Let the required distance be x km.
Then time taken during the first journey = x/2 hr.
and time taken during the second journey = x/3 hr.
x/3 + x/2 = 5 => (2x + 3x) / 6 = 5
=> 5x = 30.
=> x = 6
Required distance = 6 km.

Question 13 |

Walking 3/4

^{th}of his speed, a person is 10 sec late to his office. Find his usual time to cover the distance?40 sec | |

25 sec | |

45 sec | |

30 sec |

**Time Speed Distance**

**Discuss it**

Question 13 Explanation:

Say distance =x meters

usually he takes t sec

usual speed v = x/t m/sec ---------(1)

3v/4 = x/(t+10) ---------------(2)

(1)/(2)

=> 4/3 = (t+10)/t

=>t =30 sec

usually he takes t sec

usual speed v = x/t m/sec ---------(1)

3v/4 = x/(t+10) ---------------(2)

(1)/(2)

=> 4/3 = (t+10)/t

=>t =30 sec

Question 14 |

A boy takes 5 hours 45 min to walk from home to school come back by auto. He would have gained 2 hours by taking auto both ways. How much time would he take, to walk both ways?

6 hrs 30 min | |

7 hrs 45 min | |

7 hrs 33 min | |

6 hrs 55 min |

**Time Speed Distance**

**Discuss it**

Question 14 Explanation:

Let the distance be x km.
(Time taken to walk x km) + (Time taken to come back by auto x km)
= 5 hour 45 = 23/4 hour -----------------(1)
I both ways auto is taken time(for 2x km) = 5 hour 45 min - 2 = 15/4 hour ------(2)
(Time taken to walk 2x km)+ (Time taken to ride 2x km) = 23/2 hour => Time taken to walk 2x km=23/2−15/4
= 7 hours 45 minutes

Question 15 |

A train’s speed including stoppages, is 10 m/sec and 15 m/sec excluding stoppages. For how many seconds does the train stop per minute?

10 sec | |

15 sec | |

20 sec | |

8 sec |

**Time Speed Distance**

**Discuss it**

Question 15 Explanation:

speed of the train excluding stoppages = 15 m/sec
speed of the train including stoppages = 10 m/sec
Loss in speed when including stoppages = 15-10 = 5 m/sec
=> In 1 second, train covers 5 m less due to stoppages
Hence, time that the train stop per second = time taken to cover 5 m
=distance/speed=5/15 sec =1/3 sec
So, in 1 sec it stops for 1/3 sec

- in 60 sec it will stop for 60/3 = 20 sec.

Question 16 |

Ram and Shyam participate in a race of 200m. If Ram starts when Shyam has covered 25 meters, then Ram finishes the race 10 seconds before Shyam and if Ram starts when Shyam has covered 45 meters then both finish the race at the same time. How long does Ram take to run 200m?

120 | |

77.5 | |

87.2 | |

100 |

**Time Speed Distance**

**Race**

**Discuss it**

Question 16 Explanation:

Let time taken by Ram be Tr seconds to cover 200m.
For first case, Shyam will be covering 175m in (Tr+10)sec (As it is given that Ram takes 10sec less than Shyam)
So speed of Shyam is (175/Tr+10) m/s.
For second case, Shyam will take Tr seconds to cover 155m (200-45)
so speed of Ram is (155/Tr) m/s
Equating both speed of Ram we get 175Tr = 155(Tr+10) => Tr=77.5 s

Question 17 |

If Geeta can give a start of 100 m of distance or 20 s of time to her friend Meena in a race of 1000 meters. How much time Geeta will take to cover the 1000 meters?

200 sec | |

160 sec | |

180 sec | |

140 sec |

**Time Speed Distance**

**Race**

**Discuss it**

Question 17 Explanation:

That means Meena will cover 100 m in 20 sec

=> meena can cover 1000 m in = 200 secs

Geeta can give a start of 100 m or 20 sec to Meena

=> Geeta will take 20 sec less than Meena

=> Geeta will take 180 sec to cover 1000 meters.

=> meena can cover 1000 m in = 200 secs

Geeta can give a start of 100 m or 20 sec to Meena

=> Geeta will take 20 sec less than Meena

=> Geeta will take 180 sec to cover 1000 meters.

Question 18 |

Sita and Radha participate in a race of 1000 m. If Sita starts, when Radha has covered 100 metres, then Sita finishes the race 20 seconds before Radha, and if Sita starts, when Radha has covered 125 metres then both finish the race at the same time. How long does Sita take to run the complete race?

350 | |

700 | |

780 | |

900 |

**Time Speed Distance**

**Race**

**Discuss it**

Question 18 Explanation:

Let time taken by Sita be Ts seconds to cover 1000m.
For first case, Radha will be covering 900m is (Ts+20)sec (As it is given that Sita takes 20sec less than Radha)
So speed of Radha is (900/Ts+20) m/s
For second case, Radha will take Ts second to cover 875m (1000-125)
so speed of Radha is (875/Ts) m/s
Equating both speed of Radha we get 900Ts = 875(Ts+20) => Ts=700 sec.

Question 19 |

If Mohan can give a start of 50 m of distance or 10 s of time to Rohan in a race of 1000 meters. How much time Mohan will take to cover the 1000 meters?

190 sec | |

200 sec | |

240 sec | |

180 sec |

**Time Speed Distance**

**Race**

**Discuss it**

Question 19 Explanation:

That means Rohan takes 10 sec to cover 50 meters

=> Rohan will cover 1000 meters in 200 sec

Mohan is giving a start of 10 sec. means Mohan will take 10 sec less than Rohan to cover 1000 m.

so, Time taken by Mohan to cover 1000 m is 190 sec.

=> Rohan will cover 1000 meters in 200 sec

Mohan is giving a start of 10 sec. means Mohan will take 10 sec less than Rohan to cover 1000 m.

so, Time taken by Mohan to cover 1000 m is 190 sec.

Question 20 |

Two cars with speed of 15 kmph and 30 kmph respectively are 100 km apart and face each other. The distance between them 5 minutes before crossing is

2.75 | |

3.75 | |

4.75 | |

5.75 |

**Time Speed Distance**

**Discuss it**

Question 20 Explanation:

As the two cars are moving towards each other, their relative speed will be = 15+30 = 45 kmph.
The distance between them 5 minutes before crossing will be equal to the distance traveled by their relative speed in 5 minutes, i.e.
Required distance = Relative speed (in km per min) * time (in minute) = (45/60)*5 = 3.75 km.

There are 20 questions to complete.