# Aptitude: Time and Distance

## Formulas:

• Speed = distance/time
• To convert speed of an object from km/h to m/s, multiply it by 5 / 18.

e.g. 36 km/h = 36 x 5 /18 m/s = 10 m/s

• To convert speed of an object from m/s to km/h, multiply it by 18 / 5.

e.g. 5 m/s = 5 x 18 / 5 km/h = 18 km/h

• Relative speed :

( x + y )  km/h, if both the bodies are moving in opposite directions.

( x – y ) km/h, if both the bodies are moving in same direction.

• Train based problems:
• Time taken by a train x m long in passing a single post or pole or standing man = Time taken by the train to cover x m.
• Time taken by a train x m long in passing an object of length y m = Time taken by the train to cover ( x + y ) m.

Some shortcut formulas:

1. If A travels with speed x km/h for t1 h and with speed y km/h for t2 h, then average speed = (xt1 +  yt2) /(t1 + t2).
• If a man/vehicle covers two equal distances with the speed of x km/h and y km/h respectively, then the average speed of the man/vehicle for complete journey will be

2xy /(x + y).

• If a man changes his speed to x/y  of his usual speed and gets late by t min, then the usual time taken by him = (tx)/(y – x).if t min early then denominator becomes x-y and all is same .
• When a man travels from A to B with a speed x km/h and reaches t1 h later after the fixed time and when he travels with a speed y km/h from A to B, he reaches his destination t2 h before the fixed time, then distance between A and B

= xy(t1 + t2) /(y-x)km.

• A person travels from A to B with a speed x km/h and reaches t1 h late. Later he increases his speed by y km/h to cover the same distance and he still gets late by t2 h, then the distance between A and B =(t1 – t2)(x + y) x/y .
• If two trains start at the same time from points A and B towards each other and after crossing each other, they take t1 and t 2 time to reach points B and A respectively, then ratio of their speed = sqrt(t2/t1)

## Practice Problems:

Problem: A car completes a journey in 6 h with a speed of 50 km/h. At what speed must it travel to complete the journey in 5 h?

Solution:

Let speed be v km/h.

We know that, Distance = Speed × Time 6 x 50  = 5 × v ⇒ v = (6 x 50)/ 5 = 60 km/h

Problem: It takes 11 h for a 600 km journey if 120 km is done by train and the rest by car. It takes 40 min more if 200 km are covered by train and the rest by car. What is the ratio of speed of the car to that of the train?

Solution:

Let the speed of train be x km/h and car y km/h.

According to the question, (120 / x) + (480/ y) = 11 …(eq 1)

(200/x) + (400 / y) = 34/3  …(eq 2)

To multiply Eq. (i) by 5 and Eq. (ii) by 3, and subtracting, we get

(600 / x) + (2400 / y) – (600/ x) – (1200 / y) = 55 – 35

1200 / y = 20  ⇒ y = 60 km/h

On putting y = 60 in Eq. (i), we get

(120/x) +  (480/ 60) = 11  ⇒ (120/x)  = 11 – 8 =>  x = 40 km/hr

Required ratio = y /x = 60 /40 = 3 /2 = 3 : 2

Problem: Radha and Hema are neighbours and study in the same school. Both of them use bicycles to go to the school. Radha’s speed is 6 km/h whereas Hema’s speed is 10 km/h. Hema takes 9 min less than Radha to reach the school. How far is the school from the locality of Radha and Hema?

Solution:

Radha’s speed = 6 km/h

Hema’s speed = 10 km/h

Let the distance of the school from the locality of Radha and Hema = x km

According to the question: x / 6 – x / 10 = 9/ 60

4x = 9 =>  x = 2.25 km

Problem: A cyclist covers his first 20 km at an average speed of 40 km/h, another 10 km at an average speed of 10 km/h and the last 30 km at an average speed of 40 km/h. Then, the average speed of the entire journey is

Solution:

Time taken by cyclist to cover first 20 km = 20 / 40 = 1 /2 h

Time taken by cyclist to cover 10 km = 10 /10 = 1 h

And time taken by cyclist to cover remaining 30 km = 30 /40 = 3/ 4 h

Now, total distance travelled = (20 + 10 + 30) km = 60 km

Total time = 9/4 h Avg speed = 60 x (4/9) = 80/3 = 26.67 km/h

Problem: A car has an average speed of 60 km/h while going from Delhi to Agra and has an average speed of y km/h while returning to Delhi from Agra (by travelling the same distance). If the average speed of the car for the whole journey is 48 km/h, then what is the value of y?

Solution:

Average speed = 2ab /a + b

Here, a = 60 km/h, b = y  km/h => 48 = (2 x 60) y / (60+y)

=> 48 (60 + y )  = 120 y     => 3y = 120   =>  y = 40 km/h

Problem: In a 100 m race, A runs at 6 km/h. If A gives B a start of 8m and still beats him by 9 s, what is the speed of B?

Solution:

Speed of A = 6 km/h

Speed of A in m/s = 6 x 5 / 18 = 5/3 m/s

Time taken by A in 100 m race = (100) / (5/ 3) = 60 s

Let speed of B = x m/s

Time taken by B in (100- 8) m race = 92 / x sec

According to the question,

92 / x −  60 = 9 => 92 / x = 69

X = 92 / 69 m/s

Speed of B in km/h = 92/69 x 18/5 = 4 8. km/h

Problem: A boy went to his school at a speed of 12 km/h and returned to his house at a speed of 8 km/h. If he has taken 50 min for the whole journey, what was the total distance walked?

Solution:

Let the distance between school to house is x km.

The total time taken to cover the distance is 50 min

x /12 + x/8 = 50/60 => x (2 + 3)/24 = 50/60

x = (50 x 24) / (5 x 60)   => x = 4 km

The total distance walked by the boy is 2x = 8km

Problem: In a 100 m race A runs at a speed of 5/3 m/s. If A give a start of 4 m to B and still beats him by 12 s, what is speed of B

Solution:

Let the speed of B be x m/s.

Then, according to the question

96/x –  100×3/5 = 12

96/x –  60 = 12    =>   x = 4/3

Speed of B is 4/3 m/s

Problem: The speeds of three cars are in the ratio 2 : 3 : 4. What is the ratio between the times taken by these cars to travel the same distance?

Solution:

Let the speed of three cars be2x, 3x and 4x respectively.

Time taken by these cars to travel a distance of D are D/2x , D/3x and D/4x , respectively.

Ratio between time taken by these cars = D/2x : D/3x : D/4x

= ½ : 1/3 : ¼  => 6:4:3

Problem: A man rides one-third of the distance from A to B at the rate of x km/h and the remainder at the rate of 2 y km/h. If he had travelled at a uniform rate of 6z km/h, then he could have ridden from A to B and back again in the same time. Which one of the following is correct?

Solution:

Let the total distance be d.

Time taken to cover 1/3 rd distance,

T1 = (1/3)d / x = d/3x

Remaining distance = d – 1/3d = 2/3d

Time is taken to cover 2/3 rd distance,

T2 = (2/3)d /2y = 2d/6y

Time is taken to cover the distance from A to B and B to A, t=2d/6z

According to the question, T1 + T2 = t

⇒ d/3x + 2d/6y = 2d/6z => 1/3x + 1/3y = 1/3z

1/x + 1/y = 1/z

Problem: The distance between two points (A and B) is 110 km. X starts running from point A at a speed of 60 km/h and Y starts running from point B at a speed of 40 km/h at the same time. They meet at a point C, somewhere on the line AB. What is the ratio of AC to BC?

Solution:

Distance between two points = 110km

Their relative speed = 60 + 40 = 100 km/h

Time after which they meet = Total distance/Relative speed = 110/100 = 1.10 h

Distance covered by A in 1.10 h = AC = 60 x 1.10 = 66 km

Remaining distance = BC = 110 – 66 = 44 km Required ratio = AC: BC = 66 : 44 = 3 : 2

Problem: With a uniform speed, a car covers a distance in 8 h. Had the speed been increased by 4 km/h, the same distance could have been covered in 7 h and 30 min. What is the distance covered?

Solution:

Let the distance between A and B be x km.

Case (I)  Given, distance = x km, speed = V km/h [let] and time = 8 h

Speed = Distance/Time =  V = x/8 …(i)

Case (II) If speed =  (V + 4 )km/h and time = 15/2 h

V + 4 =  x /(15/2)  ⇒ V + 4 =  2x/15

x/8 + 4 = 2x/15  [from Eq. (i)]

x = 480 km

Problem: Travelling at 60 km/h, a person reaches his destination in a certain time. He covers 60% of his journey in 2/5 th of the time.at what speed should he travel to cover the remaining journey so that he reaches the destination right on time? (in km/h)

Solution:

We here will use the assumption let, distance = 300

Speed = 60 km/h

Time = 300/60 = 5 hrs

300 x 60% = 180 => remaining distance = 300 – 180 = 120

As it is covered in 3/5 time then speed = 120/(3/5)x5 = 40km/h

Problem: A man starts from his house and travelling at 30km/h , he reaches his office late by 10 minutes, and travelling at 24 km/h. he reaches his office late by 28 minutes .the distance (in km) from his house and office is 

Solution:

here we first will take the lcm or gcd of the times = 24 and 30 = 120

And then we will subtract the quotient of the 24 and 30 dividing 120 = 5-4 unit = 18 – 10 min

120 unit = 8/60 x 120 hr = 16 km.

Problem: To cover a distance of 416 km, a train A takes 8/3 hrs more than train B. if the speed of A is doubled, it would take 4/3 hrs less than B. what is the speed (in km/h) of train A?

Solution:

Let speed of train A = SA and speed of train B = SB

416/SA – 416/SB = 8/3

1/SA – 1/SB = 1/ 3 X 52 …(1)

416/SB – 416/2SA = 4/3

1/SB – 1/2SA = 1/ 3×104  …(2)

from eq. 1 and 2 we get SA = 52km/h

Problem: A goes to a mall from his house on a cycle at 8 km/h and comes back to his house on a cycle at 6km/h . if he takes 1 hr 10 min in all, what is the distance between his house and the mall?

Solution:

D = S1 x S2/ S1 + S2   x  total time in travelling

D = 8 x 6 / 14  x 7/6

D = 4km

Problem: A train 100 m long passes a platform 100 m long in 10 s. The speed of the train is

Solution:

Given, length of train = 100 m

Length of platform = 100 m

Time taken to pass platform = 10 s

Now, distance covered by train = 100 m +100 m = 200 m

Speed = Distance/Time   Speed = 200/10 = 20 m/s

Problem: A thief steals a car parked in a house and goes away with a speed of 40 km/h. The theft was discovered after half an hour and immediately the owner sets off in another car with a speed of 60 km/h. When will the owner meet the thief?(determine the time and distance from the house)

Solution:

Speed of thief = 40 km/h

Speed of owner = 60 km/h

Distance = Speed × Time

Distance travelled by thief in half an hour =  40 x 1/2 = 20 km

Now, relative speed of owner and thief =  (60 – 40)km/h = 20 km/h

Time taken by owner to catch thief = 20/20 = 1h

Distance covered by a thief after the theft = 40 × (1 + 1/2) = 60km The owner will meet the thief at 60 km from the owner’s house and 1.5 h after the theft.

Problem: A train moving with a speed of 60 km/h crosses an electric pole in 30 s. What is the length of the train in metres?

Solution:

Speed of train = 60 km/h

Time taken to cross pole = 30 s Distance = Speed × Time

=> Length of train = 60 x 5/18 x 30

= 500 m

Problem: A passenger train and a goods train are running in the same direction on parallel railway tracks. If the passenger train now takes three times as long to pass the goods train, as when they are running in opposite directions, then what is the ratio of the speed of the passenger train to that of the goods train? (Assume that the trains run at uniform speeds)

Solution:

Let the speed of passenger train be x km/h.

and speed of goods train be y km/h. According to the question,

x + y = 3(x – y)

x+y  = 3x – 3y

−2x  = 4 y  or  x/y = 2/1

Ratio of speed of passenger train to goods train = 2:1

Problem: A man can row at a speed of x km/h in still water. If in a stream which is flowing at a speed of y km/h it takes him z h to row to a place and back, then what is the distance between the two places?

Solution:

Downstream speed of man = (x + y) km/h

Upstream speed of man = (x – y)km /h

Let distance between two places be D km. According to the question,

D/x+y  + D/x-y  = z

D(( x – y + x + y)/x2 – y2) = z

D = z (x2 – y2)/2x

Problem: If a train crosses a km-stone in 12 s, how long will it take to cross 91 km-stones completely if its speed is 60 km/h?

Solution:

We have, speed of train = 60 km/h

Speed of train is m/s = 60 (5/18) m/s = 50/3 m/s

Train crosses a km stone in 12 s.

Length of train = speed × time = (50/3) x 12 m = 200m = 200/1000 = 1/5 km

Total distance travelled by train completely 91 km + len of train = (91 + 1/5) km = 456/5 km Time taken by train = Distance/Speed = (456)/5×60 h = 456/300 h = 1hr 30 min 12 s

Problem: A passenger train departs from Delhi at 6 pm for Mumbai. At 9 pm an express train, whose average speed exceeds that of the passenger train by 15 km/h leaves Mumbai for Delhi. Two trains meet each other mid-route. At what time do they meet, given that the distance between the cities is 1080 km?

Solution:

Let the speed of passenger train = x km/h

Speed of express train = (x + 15)km/h

Then, according to the question, 540/x  – 540/(x + 15) = 3

540 [(1/x –  1/(x + 15)] = 3

540×15 = 3x(x + 15)   =>  2700 = x2 + 15x

x2 + 15x -2700 = 0

(x+60)(x-40) = 0

x = 45

Time taken by passenger train to reach the meeting point =  540/45 = 12 h Both train will meet at (6 pm + 12 h) = 6 am

Problem: A 225 m long train is running at a speed of 30 km / hour. How much time does it take to cross a man running at 3 km / hour in the same direction?

Solution:

Speed of train = 30 km/h and speed of man = 3 km/h

Relative speed of train = Speed of train − Speed of man

= (30 – 3)km/h

= 27 km/h

= 27 (5/18) m/s = 15/2 m/s

To cross the man, train have to cover a distance equal to its length. Time to cross = Distance/Speed

= 225/ (15/2) s

= 30 s

Problem: A theif is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the theif also starts running. If the speed of the theif is 8 km / h and that of the policeman is 10 km/h, then how far will the theif have to run before he is overtaken?

Solution:

Speed of theif = 8 km/h =8 (5/18) = 20/9 m/s

Speed of policeman = 10 km/h =  10 (5/18) = 25/9 m/s

Distance between policeman and thief = 100 m

Let distance covered by theif = x m

According to the question, ((100+x)/(25/9)) = x/(20/9) 400 + 4x = 5x => x  = 400 m

Problem: A person can row downstream 20 km in 2h and upstream 4 km in 2 h. What is the speed of the current?

Solution:

Downstream speed =  20/2 = 10 km/h

Upstream speed = 4/2 = 2 km/h

Speed of current = downstream speed – upstream speed / 2 = 10 – 2 / 2 = 8/2 = 4 km/h

Problem: 27.	A train is travelling at 48 km/h completely crosses another train having half its length and travelling in opposite direction at 42 km/h in 12s. It also passes a railway platform in 45s. What is the length of the platform?

Solution:

Let the length of the train be l, then length of train travelling in opposite direction is l / 2.

Relative speed of train = (48 + 42)  = 90 km/h

Now, Distance = Speed × Time

(l + l/2 ) = 90 x 5/18 x 12

3/2 l = 25 x 12

l = 200 m

Let x be the length of the platform. As, the train crosses the platform in 45 s.

(x + l) = 48 x 5/18 x 45

x +  200 = 600  x = 400 m

Problem: A man walking at 5 km/h noticed that a 225 m long train coming in the opposite direction crossed him in 9 s. The speed of the train is

Solution:

Let the speed of the train = x km/h

Then, relative speed of train = (x + 5)km/h and length of the train = 225 m = 0 225km [given]

Time taken by train to cross the man  =  distance/time = 0.225/x+5 h

According to the question, 0.225/ (x + 5) = 9/3600  x + 5 = 90  =>  x = 85 km/h

Problem: Two trains are moving in the same direction at 1.5 km/min and 60 km/h, respectively. A man in the faster train observes that it takes 27 s to cross the slower train. The length of the slower train is

Solution:

Let the length of the slower train = x

Speed of faster train = 1.5 km/min

= 1.5×1000/60 = 25 m/s

and speed of slower train = 60 km /h =  60 (5/18) = 50/3 m/s

Since, these trains are moving in same direction.

So, relative speed of train =(25 – 50/3) m/s = 25/3 m/s

Time taken by crossing slower train = 27s

Distance = Speed × Time  x =  25/3 x 27 = 225 m

Problem: Two trains, one is of 121 m in length at the speed of 40 km/h and the other is of 99 m in length at the speed of 32 km/h are running in opposite directions. In how much time will they be completely clear from each other from the moment they meet ?

Solution:

Total length of train = 121 + 99 = 220 m

Relative speed of trains = (40 + 32)km/h = 72 km/h  i.e. 72 x 5/18  m/s = 20 m/s

Time = Distance/Speed =  220/20 = 11s

Problem: A train travels at a speed of 40 km/h and another train at a speed of 20 m/s. What is the ratio of speed of the first train to that of the second train?

Solution:

Given, speed of a train = 40 km/h =  40(5/18) m/s

Speed of another train = 20 m/s

Required ratio = Speed of first train/Speed of second train

= (40 × 5/18)/20

= 2×5 / 18

= 10/18

= 5/9 or 5:9

Problem: Normally Sarita takes 3 h to travel between two stations with a constant speed. One day her speed was reduced by 12 km/h and she took 45 min more to complete the journey. Then, the distance between the two stations is

Solution:

Applying shortcut

Let Sarita’s speed be u and distance is constant.

Then, ux3 = (u− 12) 15/4

u  = 45×4 / 3 = 60 km/h Distance = Speed × Time  = 60 x 3 = 180 km

Problem: Rani goes to school at 10 km/h and reaches the school 6 min late. Next day, she covers this distance at 12 km/h and reaches the school 9 min earlier than the scheduled time. What is the distance of her school from her house?

Solution:

Here, x = 10, y = 12, t1= 6/60  h and t2 =9/60 h

Required distance = xy(t1 + t2) / (y – x)

= 10x12x15 / 60×2

= 15 km

Problem: Two towns A B and are 250 km apart. A bus starts from A to B at 6 : 00 am at a speed of 40 km/h. At the same time another bus starts from B to A at a speed of 60 km/h. The time of their meeting is

Solution:

Let the buses meet x h after 6:00 am. Then, the distance covered by the two buses is 250 km.

40x + 60x = 250

x = 250/100

= 2.5 h

= 2h and 30 min So, they will meet at 8 : 30 am