Quantitative Aptitude Notes - Time, Speed and Distance

Quantitative Aptitude Notes - Time, Speed and Distance

Introduction:

An object that is moving at a constant rate is said to be in uniform motion and this chapter involves the calculation of the time and distance of the uniform motion object. Uniform motion problems may involve objects going in same direction, opposite directions and round trips.

Concepts:

Speed: The speed is the rate at which the body is moving (Distance travelled in a unit time). Distance covered by a body is depends on speed of the body and time taken.

Distance: The space between two objects or two points is known as distance.

Time: It is an ongoing sequence of events related to past, future and present and it is measured in seconds, minutes, hours, days, weeks, months and years.

Important Formula:

9. Two objects travel the same distance (D) with different speeds of S₁ and S₂. If the difference between the journey times is t then,

D = [(S₁ × S₂)/|S₁ – S₂|] × t

10. If an object travels a distance of D with a speed of S₁ and returns with a speed of S₂ taking a total time of 't', then,

D = [(S₁ × S₂)/|S₁ + S₂|] × t

11. If two objects travel the same distance with speeds of S₁ and S₂ taking the journey times of t₁ and t₂ respectively, then

S₁ : S₂ = t₂ : t₁ OR t₁ : t₂ = S₂ : S₁

12. Walking at p / qth of usual speed a man reaches his destination t-hours late (or early) then usual time to reach the office is given by

[(p × t)/|p – q|]

13. If a person walks a distance D with a speed of S₁ kmph then he will be late (early) by t₁ hours. If he walks at S₂ kmph he will be late (early) by t₂ hours. Then the distance travelled by the person is (M – Formula)

D = [(S₁ × S₂)/|S₁ + S₂|] × |t₁ – t₂|

Note: Use (+) sign if the person is late in one case and early in the other case. Use (-) sign if he is late in both the cases or early in both the cases.

14. If an object travels different distances with different speeds during a journey then the entire journey can be represented by a single equivalent speed called Average Speed (S_A).

Average Speed (S_A) = Total Distance (T.D) / Total Time (T. t)

15. If an object travels a distance of D₁–km with S₁–kmph another D₂–km with a speed of S₂-kmph and D₃- km with a speed of S₃-kmph, then the average speed of the journey is

S_A = {[D₁ + D₂ + D₃]/[(D₁/S₁) + (D₂/S₂) + (D₃/S₃)]}

16. If an object travels with a speed of S₁-kmph for t₁-hours, next with a speed of S₂ kmph for t₂ -hours and with a speed of S₃-kmph for t₃–hours, then the average speed (S_A) for the whole journey is

SA = [(S₁t₁ + S₂t₂ + S₃t₃)/(t₁ + t₂ + t₃)]

Memory Based Solved Example Problems based on various types

Type 1 – Basic Problems

1. The ratio between the rates of walking of A and B is 2: 3, if the time taken by B to cover a certain distance is 36 minutes, the time taken by A to cover that much distance is?

Solution:

Ratio of speed = 2: 3

Ratio of times taken for same distance = 1/2: 1/3

From question 1/2: 1/3 = t: 36

⇒ (1/3) x t = (1/2) x 36

∴ t = 54 min.

2. Sachin can cover a distance in 1hr 24 min by covering 2/3 of the distance at 4 kmph and the rest at 5 kmph. The total distance is?

Solution:

Let the total distance = D

Distance travelled at 4kmph speed = (2/3)D

Distance travelled at 5 kmph speed = (1 – 2/3)D = (1/3)D

Total time =1 hr 24 min = (60+24) min = 84/60 hr = 21/15hr

We Know, Time = Distance/Speed

Total time => 21/15 = [(2/3)D]/4 + [(1/3)D/5] => 21/15 = 2D/12 + D/15 => 21 × 4 = 10D + 4D => D = 21 × 4/14 = 6km

3. An aero plane covers a certain distance of 420 Kmph in 6 hours. To cover the same distance in 4 2/3 hours, it must travel at a speed of:

Solution:

Speed of aero plane = 420 Kmph

Distance travelled in 6 hours = 420 * 6 = 2520 Km

Speed of aero plane to cover 2520 Km in 14/3 = 2520*3/14 = 540 Km

4. A motorist travels for 6 hours, the first half at 60 Kmph and the rest at 48 Kmph. Find the distance Traveled by him.

Solution:

Distance = 3 * 60 + 3 * 48 = 180 + 144 = 324 Km

5. Laxman has to cover a distance of 6 km in 45 minutes. If he covers one half of the distance in 2/3^rd time. What should be his speed to cover the remaining distance in the remaining time?

Solution:

∵ Time left = (1/3 x 45/60) hr = 1/4 hr.

Distance left = 3km

∴ speed required = [3 / (1/4) ] km/hr = 3 x 4 =12km/hr.

6. A certain distance is covered at a certain speed. If half of the distance is covered in double time, the ratio of the two speeds is:

Solution:

Let L km distance be covered in T h. So, speed in first case = L/T km/h.

And speed in second case = (L/2)/2T = L/4T km/h

∴ Required ratio = L/T : L/4T =1 : 1/4 = 4 : 1

7. A person sets out to cycle from A to B and at the same time another person starts from B to A. After passing each other, they complete their journeys in 16 h and 25 h, respectively. Find the ratio of speeds of the 1st man to that of the 2nd man.

Solution:

Given, x = 16 and Y = 25

According to the formula,

1st man's speed: 2nd man's speed = √ y : √ x = √ 25 : √ 16 = 5 : 4

Type 2 – M – Formula

1. If I walk at 5 km/hour, I miss a train by 7 minutes. If, however, I walk at 6 km/hour, I reach the station 5 minutes before the departure of the train. The distance (in km) between my house and the station is?

Solution:

Refer to above formula (13) S1 = 5, S2 = 6, t1 = 7 min, t2 = 5 min, Td = (7+5) = 12 min  12/ 60 hr

Distance between my house and Station = 5 × 6 / | (5 -6) | × (12) /60 = 30 × 12 / 60 = 6km

2. If a man walks 20 km at 5 km/hr, he will be late by 40 minutes. If he walks at 8kmph, how early from the fixed time will he reach?

Solution:

Refer to formula (13), D = 20 km, S1 = 5, S2 = 8, t1 = 40 min, t2 = ?

20 = (5×8) / |5-8| × (40+t2) / 60 = 2 × (40 + t2) / 9 ⇒ 40 + t2 = 90 ∴t2= 50 minutes

∴The man reaches 50 minutes early from the fixed time.

3. If a train run at 40 km/hr it reaches its destination late by 11 min but if it run at 50 km/hr it is late by 5 min only. The correct time for the train to cover at journey is?

Solution:

Let the required time = T min.

Then, distance covered in T + 11 min at 40 km/hr = distance covered in T + 5 min at 50 km/hr.

⇒ 40 (T + 11) / 60 = 50 (T + 5) / 60

∴ T = 19 min.

Type 3 – a/b => Speed

1. Walking 6/7th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover the distance is?

Solution:

Refer to formula (12) above, p/q = 6/7, t = 12

Usual time = 6 × 12 / |6-7| = 72 min = 1 hr 12 min

2. A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?

Solution:

3. A car travels a distance of 840 km at uniform speed. If the speed of the car is 10 km/hr more it takes two hours less to cover the same distance. The original speed of the car was?

Solution:

Let the original speed be x km/hr.

Then, [840/x - 840/( x + 10) ] = 2

⇒ 840 (x + 10) - 840x = 2x (x+10)

⇒ x2 + 10x - 4200 = 0

⇒ (x + 70) (x - 60) = 0

∴ x = 60 km/hr.

4. A train running at 8/11 of its own speed reached a place in 5 1/2 h. How much time could be saved, if the train would have run as its own speed?

Solution:

New speed = 8/11 of usual speed

New time = 11/8 of usual time

∴ 11/8 of usual time = 11/2 h

⇒ Usual time = (11 x 8)/(2 x 11) = 4 h

∴ Time saved = 5 1/2- 4 = 1 ½

Type 4 – Average Speed

1. If a person travel 39 km at a speed of 26 km/h, another 39 km at a speed of 39 km/h and 39 km again at a speed of 52 km/h . Then find the average speed of entire journey

Solution:

Average speed of entire journey = (39 + 39 + 39)/(39/26 + 39/39 +39/52) = 3/(1/26 + 1/39 +1/52)

= 3/[(6 + 4 + 3)/156]

= 3 x 156/13 = 36 km/h

2. One third of a certain journey is covered at the rate of 25km/hour, one-fourth at the rate of 30 km/hour and the rest of 50 km/hour. The average speed for the whole journey is?

Solution:

Let the total distance covered = 600 km

Distance covered with the speed of 25 km/hr = 200 km

Distance covered with the speed of 30 km/hr = 150 km

Distance covered with the speed of 50 km/hr = 250 km

Total time taken to cover the distance = 200/25 + 150/30 + 250/50 = 8+5+5=18 h

Average speed = 600/18 = 33 1/3 km/hr

3. The average speed of a bus is 55 kmph. How much distance will a car cover in eight hours if the average speed of the car is 26 percent more than the average speed of the bus?

Solution:

Sb = 55 kmph, Sc = 55×126/100 = 69.3, t = 8 h

D =S× t = 69.3 × 8 = 554.4 km

4. The average speed of a train is 1 3/7 times the average speed of a car. The car covers a distance of 588 km in 6 hours. How much distance will the train cover in 13 hours?

Solution:

Speed of car = 588/6 = 98

Speed of train =10/7 * 98 = 140 kmph

Distance covered by the train = 13 * 140 = 1820 km

5. A car reached Raipur from sonagarh in 35 min with an average speed of 69 km/h. If the average speed is increased by 36 km/h, how long will it take to cover the same distance?

Solution:

Distance between Sonagarh and Raipur = Average speed x time = 69 x (35/60) km = 161/4 km

New speed = (69 + 36) km/h = 105 km/h

∴ Required time = Distance/speed = 161/(4 x 105) h = (161 x 60)/(4 x 105) min = 23 min

Type 5 – Based on Wheel

1. A car is moving with the speed of 47.52km/hr and the radius of the wheel of car is 21 cm. Calculate the approximate number of revolutions made by the wheel in one minute.

Solution:

Speed in cm/minute = (Speed in km/hr x 1000 x100 )/60 = 47.52x(50/3) =79200 cm/min

And Circumference of circle = 2πr = 2 x (22 / 7) x 21 =132

No. of revolutions = (Speed in cm/minute) / circumference of circle in cm =79200 / 132 =600 rpm

2. The wheel of an engine of 300 cm in circumference makes 10 revolution in 6 seconds. what is the speed of the wheel (in km/h) ?

Solution:

Circumferences means one resolutions.

Therefore, distance covered in 10 resolutions =300 x 10 = 30 m i.e., 30 meters in 6 seconds.

∴ Speed of wheel = 30/6 m/s = 5 m/s

∴ 5 m/s = 5 x (18/5) = 18 km/h

Type 6 – Walking and Riding

1. Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.

Solution:

Distance = Speed × time

Here time = 2hr 45 min = 11/4

Distance = 4 x 11/4 = 11km

New Speed = 16.5km

Therefore time = 11/16.5 = 40 min

2. A man travelled from the village to the post – office at the rate of 20 kmph and walked back at the rate of 5 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post – office from the village?

Solution:

Let, x = 20 kmph, y = 5 kmph

Given, time took for the whole journey = 5 hours 48 minutes

Now, Average speed = 2xy/(x+y)

⇒ Average speed = 2∗20∗5/(20+5)

⇒ Average speed = 200/25

⇒ Average speed = 8 kmph

So, Distance travelled in 5 hours 48 minutes i.e 5 x 4/5 hours = 8 x 29/5 = 46.4 km

Therefore, Distance between the post – office from the village = 46.4/2 = 23.2 km

3. If a man decides to travel 80 km in 8 hours, partly by foot and partly on, a bicycle, his speed on foot being 8km/hr and that on bicycle being 16 km/hr, what distance would he travel on foot?

Solution:

Let the time taken to travel a part of distance by foot = x

∴Time taken to travel by bicycle = 8–x

8 × x + 16 (8 –x) = 80

8x + 128 – 16x = 80

8x = 48 => x = 48 / 8 = 6

Distance travelled by foot = 8 × 6 =48 km

4. A man takes 4 h 20 min in walking to a certain place and riding back. If he walk on both sides he losses 1 h. The time he would take by riding both ways is?

Solution:

W + R = 4 h 20 min

W + W = 5 h 20 min

∴ R + R = 3 h 20 min

Type 7 – Relative Speed

1. A thief is noticed by a policeman from a distance of 300 Metres. The thief starts running and policeman chases him. The thief and the policeman run at the rate of 9 Kmph and 12 Kmph respectively. What is the difference between them after 3 Minutes?

Solution:

Speed of police man = 12 Kmph

and speed of thief = 9 Kmph

their relation speed = (12 - 9) = 3 Kmph

Distance travelled in 1/20 Hour

Thus, the distance between them = 150 Metres.

2. A thief sees a jeep at a distance of 250 m, coming towards him at 36 km/h. Thief takes 5 seconds to realize that there is nothing but the police is approaching him by the jeep and start running away from police at 54 km/h. But police realize after 10 seconds, when the thief starts running away, that he is actually a thief and gives chase at 72 km/h. how long after thief saw police did police catch up with him and what is the distance police had to travel to do so?

Solution:

Initial speed of police = 10 m/s

Increased speed of police = 20 m/s

Speed of thief = 15 m/s

Initial difference between thief and police = 250 m

After 5 seconds difference between thief and police = 250 - (5 x 10) = 200 m

After 10 seconds more, the difference between thief and police = 200 + (5 x 10) = 250 m.

Now, the time required by police to catch the thief = 250/5 = 50 s

Distance travelled = 50 x 20 = 1000 m

Total time = 50 + 15 = 65 sec

Total distance = 1000 + (15 x 10) = 1150 m

3. A walks at a uniform rate of 4 km an hour and 4 hours after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A?

Solution:

Distance covered by A in 4 hours = 16 km

For every hour B covers 6 km more than A.

So he covers 16 km in 16/6 = 8/3 hours

Distance covered by B in hours is 10 * 8/3 = 80/3= 26.7 km

4. The ratio between the rates of travelling of A and B is 2:3 and therefore A takes 10 min more than the time taken by B to reach a destination, if A had walked at double the speed, he would have covered the distance in?

Solution:

Ratio of times taken by A and B = 1/2 : 1/3

Suppose B takes Tb min. Then A takes (Tb + 10) min.

∴ (Tb + 10) : Tb = 1/2: 1/3

⇒ (Tb + 10)/Tb = 3 / 2

⇒ 2Tb + 20 = 3Tb => ∴ Tb = 20

∴ Time taken by A = 20 +10 = 30 minute

If A had walked at double speed

Req. time = 30/2 =15 minute

5. Two boys begin together to write out a booklet containing 817 lines. The first boy start with first line, writing at rate of 200 lines an hour and the second boys start with a last line. He writes line 817 and so on backwards proceeding at the rate of 150 lines an hour. An what line will they meet?

Solution:

Let the two meet at the Nth line

From the question

Nth/200 = (817 - Nth)/150

⇒ 3Nth = 4(817-Nth)

⇒ Nth = (4 x 817) / 7

∴ Nth = 466.85

So, they will meet at the 467th line

6. Den Bosh and Eastbourne are two famous cities 300 km apart. Maradona starts from Den Bosch at 8 : 24 am. An hour later Pele starts from Den Bosch. After travelling for 1 hour, Pele reaches Nottingham that Maradona had passed 40 minutes earlier. Nottingham falls on the way from Den Bosh to Eastbourne at the same time, what is the speeds of the time Maradona and Pele respectively?

Solution:

Let pele covers D km in 1 hours. So Maradona takes (2 h - 40 min) = 1 h 20 min to cover D km.

Let speed of Maradona and pele be M and P respectively than.

D = M x 4/3 and D = P x 1

M/P = 3/4

Again, 300/M - 300/P = 1

300/3k - 300/4k = 1 => k = 25

M = 3k = 75 km/h and

P = 4k = 100 km/h

Type 8 – Miscellaneous

1. The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms, then what was the total distance travelled in 12 hours?

Solution:

Total distance travelled in 12 hours = (35+37+39+.....upto 12 terms)

This is an A.P with first term, a=35, number of terms, n= 12,d=2.

Required distance = 12/2[2 x 35+{12-1) x 2]=6(70+23)= 552 kms.

2. Excluding stoppages, the speed of a bus is 54 km/hr. and including stoppages it is 45 km/hr. for how many minutes does the bus stop per hour?

Solution:

Due to stoppages, it covers 9 km less per hour

Required time = Time taken to cover 9 km = 9 / (54 /60) min = 10 min

3. For every 4 leaps taken by a dog, a cat takes 5 leaps in the same time. But distance covered in 3 leaps of dog is same as that of 4 leaps of cat. Then what is the ratio of speed of dog and cat?

Solution:

Let distance covered by dog in 1 leap is x and

Distance covered by cat in 1 leap is y

Then, 3x = 4y => ∴ x= 4/3y

Now, Ratio of speed of dog and cat = Ratio of distance covered by them in the same time = 4x : 5y = 16/3y: 5y => 16: 15

4. Suresh traveled 1200 km by air which formed (2/5) of his trip. One-third of the whole trip he traveled by car And the rest of the journey he performed by train. The distance traveled by train was?

Solution:

Let the total distance be y km.

Then, 2y / 5 = 1200

⇒ y = (1200 x 5) /2 = 3000 km.

Distance traveled by car = (1/3 X 3000) = 1000 km.

Distance traveled by train = [3000- (1200 + 1000) ] km. = 800 km.

5. A start from P walk to Q a distance of 51.75 kilometer at the rate of 3.75 km an hour. An hour latter B starts from Q for P and walk at the rate of 4.25 km an hour. When and where will A meet B?

Solution:

A has already gone 3.75 km when B starts of the remaining 48 km.

A walks 3.75 km and B walks 4.25 km in one hour in opposite direction

i.e., they together pass over (3.75 + 4.25) = 8 km in hour.

Therefore, 48 km. are passed over in 48/8= 6 hours.

∴ A meets B in 6 hours after B started and, therefore, they meet at a distance of (4.25 × 6 ) = 25.5km from Q.

6. A person leaves a place A to place B at 6 a.m. and reaches place B at 10 a.m. Another person leaves B at 7:30 a.m. and reaches A at 11 a.m. They will meet each other at?

Solution:

Let distance be 28 km

Speed of the first person = 28/4 = 7

Speed of the second person = 28/3.5 = 8

Distance covered by the first person upto 7:30 a.m. = 7 × 1.5 = 10.5

Remaining distance = 28–10.5 = 17.5 km

Time taken by both the persons to travel 17.5 km = 17.5 / 15 = 1 hr 10 min

Time they met = 7:30 + 1 hr 10 min = 8:40 a.m.

7. Soniya and Priyanka started from Amethi and Bellari for Bellari and Amethi, which are 645 km apart. They meet after 15 hours. After their meeting, Sonia increased her speed by 3 km/h and priyanka reduced her speed by 3 km/h, they arrived at Bellari and Amethi respectively at the same time. What is their initial speeds?

Solution:

The sum of their speeds = 615/15 = 43 km/h

Notice that they are actually exchanging their speeds. Only then they can arrive at the same time at their

respective destinations. It means the difference in speeds is 3 km/h.

Thus, x + ( x + 3) = 43

x = 20 and x + 3 = 23

The concept is very similar to the case when after meeting each other they returned to their own places of departure. It can be solved through option also.

8. The ratio of speeds of a train and a car is 16 : 15 , respectively and a bus covered a distance of 480 km in 8h. The speed of the bus is 3/4th of the speed of train. What will be covered by car in 6 h?

Solution:

Let speed of train = 16k and Speed of car = 15k

Speed of bus = 480/8 = 60 km/h

According to the question,

16k = 60 x 4/3 = 80 ⇒ k = 5

∴ Speed of car = 15k = 15 x 5 = 75 km/h

∴ Required distance = Speed x Time = 75 x 6 = 450 km