# PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

If a man walks at the rate of 5 km/h, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/h, he reaches the station 8 minutes before the arrival of the train. Find the distance covered by him to reach the station.

- (a)
4 kms

- (b)
7.5 kms

- (c)
5 kms

- (d)
7 kms

Answer: Option B

**Explanation** :

Let the required distance be x km.

Difference in the times taken at two speeds = 15 mins = 1/4 hrs.

∴ x/5 - x/6 = 1/4

⇒ 6x - 5x = 7.5

⇒ x = 7.5

Hence, the required distance is 7.5 kms.

Hence, option (b).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

In a 1 km race, Varun takes 3 minutes more than Gautam and 1 minute less than Tarun. If Gautam runs at a speed of 10 km/h in 1 km race, what is Tarun's speed?

- (a)
9 kmph

- (b)
7 kmph

- (c)
8 kmph

- (d)
6 kmph

- (e)
5 kmph

Answer: Option D

**Explanation** :

Gautam covers 1 km at a speed of 10 km/h.

So, he takes 6 minutes to cover the distance.

Varun will take (6 + 3), i.e. 9 minutes and Tarun will take 9 + 1 = 10 minutes to cover the same distance.

∴ Tarun's speed is 1/(10/60) km/h = 6 km/h.

Hence, option (d).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

The speed of a car increases by 2 kms after every one hour. If the distance travelled in the first one hour was 10 kms, what was the total distance travelled in 12 hours?

- (a)
256 kms

- (b)
282 kms

- (c)
252 kms

- (d)
None of these

Answer: Option C

**Explanation** :

Total distance travelled in 12 hours = (10 + 12 + 14 + …. upto 12 terms)

This is an A.P. with first term, a = 10, number of terms, n = 12, common difference d = 2.

∴ Required distance =12/2 × [2 × 10 + (12 - 1) × 2] = 6 × (20 + 22) = 252 km

Hence, option (c).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

It takes eight hours for a 500 kms journey if 300 kms is done by train and the rest by car. It takes 60 minutes lesser if 300 kms is done by car and the rest by train. The ratio of the speed of the train to that of the speed of the car is

- (a)
4 : 3

- (b)
3 : 4

- (c)
1 : 2

- (d)
2 : 1

Answer: Option C

**Explanation** :

Let the speed of the train and the car be t km/h and c km/h, respectively.

Now, $\frac{300}{t}+\frac{200}{c}$ = 8 …(1)

and $\frac{300}{c}+\frac{200}{t}$ = 7 …(2)

From (1), 300c + 200t = 8ct and …(3)

From (2), 300t + 200c = 7ct …(4)

From (3) and (4), eliminating ct term we get

t : c = 1 : 2

Hence, option (c).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A person has to cover a distance of 6 km in 45 minutes, if he covers one half of the distance in two – thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/h) must be:

- (a)
6

- (b)
8

- (c)
12

- (d)
15

Answer: Option C

**Explanation** :

Remaining distance = 3 km and remaining time = (1/3 × 45) mins = 15 mins = 1/4 hour.

∴ Required speed =(3 × 4) km/hr = 12 km/hr

Hence, option (c).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

An aero plane first flew with a speed of 330 km/h and covered a certain distance. It still had to cover 110 km less than what it had already covered, but it flew with a speed of 440 km/h. The average speed for the entire flight was 374 km/h. Find the total distance covered.

- (a)
1850 kms

- (b)
1870 kms

- (c)
2200 kms

- (d)
1950 kms

Answer: Option B

**Explanation** :

Let the aero plane cover x km at a speed of 330 km/h and (x – 110) km at a speed of 440 km/h.

Hence, it covers a total distances of (2x – 110) km at a speed of 374 km/h

Average speed = (Total Distance)/(Total Time)

⇒ 374 = $\frac{2x-110}{\frac{x}{330}+\frac{x-110}{440}}$

or $\frac{2x-110}{374}=\frac{x}{330}+\frac{x-110}{440}$

or x = 990

Therefore, the total distance covered = 2x-110=2×990-110=1870 km

Hence, option (b).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A train starts from Agra at 6:00 AM and reaches Delhi at 10:00 AM. The other train starts from Delhi at 8:00 AM and reaches Agra at 11:30 AM. If the distance between Agra and Delhi is 200 km, then what time did the two trains meet each other?

- (a)
8:56 a.m.

- (b)
8:46 a.m.

- (c)
7:56 a.m.

- (d)
8:30 a.m.

Answer: Option A

**Explanation** :

Speed of the first train = 50 km/hr

Speed of second train = $\frac{400}{7}$ km/hr

At 8:00 AM distance between two trains is 100 kms.

Relative velocity = 50 + $\frac{400}{7}$ = $\frac{350+400}{7}$ = $\frac{750}{7}$ km/h

Time taken = $\frac{100\times 7}{750}$ × 60 = 56 mins.

Hence, the two trains meet each other at 8:56 am.

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Chahar can row a certain distance downstream in 6 hours and return the same distance in 9 hours. If the speed of Chahar in still water is 12 km/h, find the speed of the stream.

- (a)
2 kmph

- (b)
2.4 kmph

- (c)
3 kmph

- (d)
Cannot be determined

Answer: Option B

**Explanation** :

Let the speed of the stream be x km/hr and distance travelled be S km. Then,

$\frac{S}{12+x}$ = 6 and $\frac{S}{12-x}$ = 9

⇒ $\frac{12-x}{12+x}=\frac{6}{9}$

⇒ 108 - 9x = 72 + 6x

⇒ 15x = 36

⇒ x = 36/15 = 2.4 km/hr

Hence, option (b).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A man sitting in a train travelling at the rate of 50 km/h observes that it takes 9 sec for a goods train travelling in the opposite direction to pass him. If the goods train is 187.5 m long, find its speed.

- (a)
40 kmph

- (b)
25 kmph

- (c)
35 kmph

- (d)
36 kmph

Answer: Option B

**Explanation** :

Distance covered = 187.5m, Time = 9 secs

Relative speed = $\frac{187.9}{9}\times \frac{3600}{1000}$ = 75 km/hr

As the trains are travelling in opposite directions, speed of goods train = 75 – 50 = 25 km/hr

Hence, option (b).

Workspace:

**PE 1 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Piyush can run a km in 8 minutes and Kishan in 8 minutes 20 seconds. How many meters start should Piyush give Kishan so that the race may end in a dead heat?

- (a)
40 meters

- (b)
60 meters

- (c)
80 meters

- (d)
None of these

Answer: Option A

**Explanation** :

Piyush runs a km in 8 mins = 480 secs and Kishan runs a km in 500 secs.

∴ Piyush beats Kishan by 20 secs.

So, in order to end the race in a dead heat, Piyush should give Kishan a start of ‘x’ m or 20 secs for Kishan,

In 500 secs Kishan runs 1000 m.

∴ In 20 secs Kishan will run 1000/500 × 20 = 40 m

Hence, option (a).

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.