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Time, Speed and Distance Exercise 2

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Time, Speed and Distance 2

  • This online quiz will test your knowledge of Time, Speed and Distance in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

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

Total distance travelled in 12 hour

= (35 + 37 + 39 + ..... upto 12 terms)

This is an A.P. with first term a = 35,

number of terms n = 12,

common difference d = 2.

∴ Required distance

= time-and-distance-q-67767.png

Anna left for city A from city B at 5.20 a.m. She travelled at the speed of 80 km/h for 2 hours 15 minutes. After that the speed was reduced to 60 km/h. If the distance between two cities is 350 kms, at what time did Anna reach city A?

Distance covered in 2 hr 15 min,

i.e., 2¼ hrtime-and-distance-q-67737.png

Time taken to cover remaining distance

= time-and-distance-q-67731.png

= 2 hr 50 min.

Total time taken

= (2 hr 15 min + 2 hr 50 min)

= 5 hr 5 min.

So, Anna reached city A at 10.25 a.m.

A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9 : 30 a.m.?

Relative speed

= (2 + 3) = 5 rounds per hour.

So, they cross each other 5 times in an hour and 2 times in half an hour.

Hence, they cross each other 7 times before 9 :.30 a.m.

A circular running path is 726 metres in circumference. Two men start from the same point and walk in opposite directions @ 3.75 km/h and 4.5 km/h, respectively. When will they meet for the first time?

Their relative speeds

= (4.5 + 3.75) = 8.25 km/h

Distance = 726 metres

= time-and-distance-q-68030.png

Required time

= time-and-distance-q-68024.png min

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:

Remaining distance = 3 km

and remaining time

= time-and-distance-q-68018.png

∴ Required speed = (3 × 4) km/h

= 12 km / hr.

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

Let the duration of the flight be x hour Then,

time-and-distance-q-68012.png

⇒ x (2x + 1) = 3

⇒ 2x² + x – 3 = 0

⇒ (2x + 3) (x – 1) = 0

⇒ x = 1 hr. [neglecting the –ve value of x].

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

Relative speed of the thief and policeman

= (11 – 10) km/h = 1 km/h.

Distance covered in 6 minutes

= time-and-distance-q-68006.png

∴ Distance between the thief and policeman

= (200 – 100) m = 100 m.

Two persons living in different towns 32 km apart started at 1 pm to see each other at the rate of 3.5 and 4.5 km per hour respectively. When will they meet?

Relative speed

= (3.5 + 4.5) = 8 kmph.

Time to meet = 32/8 = 4 hour

So when they meet at 5 pm, one will have walked

3.5 × 4 = 14 km and the other will have walked

4.5 × 4 = 18 km.

A man rides at the rate of 11 miles an hour, but stops for 5 min to change horse at the end of every seventh mile. How long will be take to go a distance of 96 miles? (Approx.)

time-and-distance-q-67999.png

Time taken to travel 96 miles

= ⁹⁶⁄₁₁ hr = 8 hr 43 minutes

During the journey of 96 miles, he has to stop for 13 times to change the horse.

∴Total stoppage time

= 13 × 5 = 65 mins. = 1 hr. 5 mins.

Hence the total time

= 8 hr 43 mins + 1 hr. 5 mins.

= 9 hr 48 mins.

Subbu starts from a point O at 10:00 a.m., overtakes Ajay, who is moving in the same direction, at 11:00 a.m. and Bhuvan moving in the opposite direction at 12:00 (noon). If the speed of Bhuvan is one fourth the speed of Subbu, at what time will Ajay and Bhuvan cross each other?

Let the speed of Ajay be V and the speed of Bhuvan and Subbu be 1 and 4 respectively.

Then OA = 4 and OB = 4.

At 12:00 noon.

time-and-distance-q-67992.png

Let Ajay be at C at 12:00 noon at a distance of V from A (towards B)

∴ Time taken for them to meet from 12:00 noon. = time-and-distance-q-67986.png

Since V is not known time-and-distance-q-67979.pngcannot be determined.

On a journey across Bombay, a tourist bus averages 10 km/h for 20% of the distance, 30 km/h for 60% of it and 20 km/h for the remainder. The average speed for the whole journey was

Let the average speed be x km/h.

and total distance = y km. then,

time-and-distance-q-68400.png

time-and-distance-q-68394.png

A man walks half of the journey at 4 km/h by cycle does one third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hours and 12 minutes. The length of the journey is

Let the length of the journey =x km.

∴ Journey rides by horse cart

time-and-distance-q-68387.pngtime-and-distance-q-68381.png

Then, total time taken to complete journey

time-and-distance-q-68375.png

time-and-distance-q-68369.png

time-and-distance-q-68362.png

time-and-distance-q-68355.png

A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10: 30 a.m. At what time do the two trains cross each other?

Let the distance between X and Y be x km.

Then, the speed of A is x/4 km/h and that of B is 2x/7 km/h.

time-and-distance-q-68337.png

Relative speeds of the trains

= time-and-distance-q-68331.png

Therefore the distance between the trains at 7 a.m.

= time-and-distance-q-68324.png

Hence, time taken to cross each other

= time-and-distance-q-68318.png

Thus, both of them meet at 7 : 56 a.m.

A man covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 minutes less. If it moved 2 km/h slower, it would have taken 20 minutes more. Find the distance.

Let the distance be x km. Let speed of train be y km/h. Then by question, we have

time-and-distance-q-68312.png ...(i)

and time-and-distance-q-68305.png ...(ii)

On solving (i) and (ii), we get x = 3y

Put x = 3y in (i) we get

time-and-distance-q-68299.png

⇒ y = 20

Hence, distance = 20 × 3 = 60 km.

If a man walks to his office at 5/4 of his usual rate, he reaches office 30 minutes early than usual. What is his usual time to reach office.

time-and-distance-q-68293.png

time-and-distance-q-68287.png

Mary jogs 9 km at a speed of 6 km per hour. At what speed would she need to jog during the next 1.5 hours to have an average of 9 km per hour for the entire jogging session?

Let speed of jogging be x km/h.

Total time taken

time-and-distance-q-68281.png

Total distance covered = (9 + 1.5x) km.

time-and-distance-q-68275.png

time-and-distance-q-68269.png

Mohan travels 760 km to his home, partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if he travels 240 km by train and the rest by car. The speed of the train and the car, respectively are:

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

Now, time-and-distance-q-68263.png …(i)

and time-and-distance-q-68257.png …(ii)

Solving (i) and (ii),

we have x = 80 km/h and

y = 100 km/h.

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the speed of the car is :

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

Now, time-and-distance-q-68251.png …(i)

and time-and-distance-q-68245.png …(ii)

From (i),

120y + 480x = 8xy and …(iii)

From (ii),

200y + 400x = time-and-distance-q-68239.png …(iv)

From (iii) and (iv),

time-and-distance-q-68233.png

⇒ 15y + 60x = 24y + 48x

⇒ 12x = 9y

⇒ time-and-distance-q-68226.png

In a flight of 6000 km, an aircraft was slowed down due to bad weather. The average speed for the trip was reduced by 400 kmph and the time of flight increased by 30 minutes. The original planned duration of the flight was

Let the original planned time of the flight be x hour

∴ The average speed of the flight

= time-and-distance-q-68220.png km/h

If the average speed is

time-and-distance-q-68214.png km/h,

then the time of the flight is

time-and-distance-q-68208.png hour

∴ time-and-distance-q-68202.png

⇒ – 400x + time-and-distance-q-68196.png

⇒ – 4x + time-and-distance-q-68190.png

⇒ – 4x² + 30 – 2x = 0

⇒ 4x² + 2x – 30 = 0

⇒ 2x² + x – 15 = 0

⇒ x = time-and-distance-q-68184.png

= time-and-distance-q-68178.png

Points A and B are 70 km apart on a highway. One car starts form A and the another one from B at the same time. If they travel in the same direction, they meet in 7 hours But if they travel towards each other, they meet in one hour. The speeds of the two cars are, respectively.

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

Their relative speeds when they are moving in same direction = (x – y) km/h.

Their relative speeds when they are in opposite directions = (x + y) km/h.

Now, time-and-distance-q-68172.png

or x + y = 70 ... (i)

and time-and-distance-q-68166.png or x – y = 10 ... (ii)

Solving (i) and (ii), we have

x = 40 km/h and y = 30 km/h.

A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/h from the usual speed. Find its usual speed.

Let the usual speed of the plane be x km/h

∴ Time taken in covering the distance of 1500 km

=1500/x hour

∴ time-and-distance-q-68160.png

⇒ 3000 x = 3000 (x + 250) – x(x + 250)

⇒ x² + 250 x – 3000 × 250 = 0

⇒ time-and-distance-q-68153.png

= time-and-distance-q-68147.pngkm/h

Walking ⁶⁄₅ of his usual speed, a person takes 10 min less to reach his office. His usual time taken to reach the office is

time-and-distance-q-68141.png

A man starts from B to K, another from K to B at the same time. After passing each other they complete their journeys in 3⅓ and 4⅘ hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.

time-and-distance-q-68135.png

= time-and-distance-q-68129.png

= time-and-distance-q-68123.png

= time-and-distance-q-68117.png

= time-and-distance-q-68111.png = ⁶⁄₅

∴ time-and-distance-q-68104.png = ⁶⁄₅

∴ 2nd man’s speed

= ⁶⁰⁄₆ = 10 km/hr.

A man travelled a distance of 50 km on his first trip. On a later trip, travelling three times as fast, he covered 300 km. Compare the times he took.

Let speed be x

⇒50/x = time taken

also time-and-distance-q-68098.png = time taken.

Hence ratio is 1 : 2.

The driver of a car driving @ 36 kmph locates a bus 40 meters ahead of him. After 20 seconds the bus is 60 meters behind. The speed of the bus is :

Net distance gained by car over the bus

= 40 + 60 = 100m, in 20 sec.

Time = time-and-distance-q-68092.png

⇒ S₂ = 5 m/s = 18 kmph.

Now check your Result..

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