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All questions of Time, Speed and Distance for Mechanical Engineering Exam

Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
  • a)
    17 hr
  • b)
    14 hr
  • c)
    12 hr
  • d)
    19 hr
Correct answer is option 'A'. Can you explain this answer?

In this type of questions we need to get the relative speed between them, 
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
  • a)
    80 km
  • b)
    70 km
  • c)
    60 km
  • d)
    50 km
Correct answer is option 'D'. Can you explain this answer?

Alok Verma answered
Distance he could travelled/speed diff.
= 20/(14-10)
= 20/4
= 5 hrs
Now his actual speed was 10 km/h
Total distance travelled by him = speed × time
= 10 × 5
= 50 km.
 

A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
  • a)
    12km
  • b)
    14km
  • c)
    16km
  • d)
    18km
Correct answer is option 'C'. Can you explain this answer?

Arun Sharma answered
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

So distance traveled on foot = 4(4) = 16 km

A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?
  • a)
    5
  • b)
    6
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)
=2+3 = 5 rounds per hour
Therefore, they cross each other 5 times in 1 hour and 2 times in 1/2 hour
Time duration from 8 a.m. to 9.30 a.m. = 1.5 hour
Hence they cross each other 7 times before 9.30 a.m.

A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
  • a)
    121 km
  • b)
    242 km
  • c)
    224 km
  • d)
    112 km
Correct answer is option 'C'. Can you explain this answer?

Dhruv Mehra answered
Let time taken to travel the first half = x hr 
Then time taken to travel the second half = (10 - x) hr 

Distance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 - x)

Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 - x)
=> 45x = 240
=> x = 16/3
Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]

Practice Quiz or MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for Time & Distance under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
 
Q. A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
  • a)
    11 hrs
  • b)
    8 hrs 45 min
  • c)
    7 hrs 45 min
  • d)
    9 hrs 20 min
Correct answer is option 'C'. Can you explain this answer?

Manoj Ghosh answered
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
In fact, you can do all these calculations mentally and save a lot of time
which will be a real benefit for you.

A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?
  • a)
    600 m
  • b)
    0.6 m
  • c)
    6 km
  • d)
    900 m
Correct answer is option 'A'. Can you explain this answer?

Rajeev Kumar answered
Lalloo is unfortunate that the friend is moving away from him.
(Because the friend moves in same direction as Lalloo).
relative speed= 20- 15= 5,kmph. distance= 200 m.
Thus, Lalloo will meet his friend when he gains 200 m over him.
=> time required = distance / speed = 0.2/5 = 1/25 hrs.
=> Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)
=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m

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. What is the ratio of the speed of the train to that of the car?
  • a)
    3 : 4
  • b)
    2 : 3
  • c)
    1 : 2
  • d)
    1 : 3
Correct answer is option 'A'. Can you explain this answer?

Arya Roy answered
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y=8      1/x + 4/y = 1/15 ...(i)
And, 200/x + 400/y = 25/3  1/x + 2/y = 1/24   ...(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.

Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
  • a)
    1 hr 42 min
  • b)
    1 hr
  • c)
    2 hr
  • d)
    1 hr 12 min
Correct answer is option 'D'. Can you explain this answer?

Kavya Sharma answered
New speed = (6/7) of usual speed.
New time = (7/6) of usual time.
Therefore (7/6 of usual time)- (usual time) = (1/5) hr.
=> (1/6 of usual time)= (1/5) hr 
=> usual time = (6/5) hr 
= 1 hr 12 min.

A person going from Pondicherry to Ootacamond travels 120 km by steamer, 450 km by rail and 60 km by horse transit. The journey occupies 13 hours 30 minutes, and the speed of the train is three times that of the horse-transit and 1(1/2) times that of the steamer. Find the speed of the train.
  • a)
    20 kmph
  • b)
    60 kmph
  • c)
    10 kmph
  • d)
    50 kmph
Correct answer is option 'B'. Can you explain this answer?

Aarav Sharma answered
Given data:
Total distance travelled = 120 km + 450 km + 60 km = 630 km
Total time taken = 13 hours 30 minutes = 13.5 hours

Let the speed of the steamer be x kmph.
Then, the speed of the horse transit = x/1.5 = 2x/3 kmph (as given, the speed of the train is 1.5 times that of the steamer)
And, the speed of the train = 2x kmph (as given, the speed of the train is three times that of the horse-transit)

Calculation:
Let's assume the time taken by the steamer, train, and horse transit are t1, t2, and t3 respectively.
Then, we have:
t1 + t2 + t3 = 13.5 hours - - - (1) (Total time taken)
t1 = 120/x - - - (2) (Time taken by steamer = Distance/Speed)
t2 = 450/2x - - - (3) (Time taken by train = Distance/Speed)
t3 = 60/(2x/3) = 90/x - - - (4) (Time taken by horse transit = Distance/Speed)

Substituting the values of t1, t2, and t3 in equation (1), we get:
120/x + 450/2x + 90/x = 13.5
Simplifying this equation, we get:
x = 60 kmph

Therefore, the speed of the train is 2x = 120 kmph.
Hence, the correct option is (b) 60 kmph.

In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun's speed?
  • a)
    8 kmph
  • b)
    5 kmph
  • c)
    4 kmph
  • d)
    7 kmph
Correct answer is option 'B'. Can you explain this answer?

If Arun doubles his speed, he needs 3 hour less. Double speed means half time. Hence, half of the time required by Arun to cover 30 km = 3 hours

i.e., Time required by Arun to cover 30 km = 6 hours

Arun's speed = 30/6 = 5 km/h 

Two men A and B walk from P to Q, a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. The distance from P to R is
  • a)
    12 km
  • b)
    18 km
  • c)
    10 km
  • d)
    24 km
Correct answer is option 'B'. Can you explain this answer?

Solution:

Let's break down the problem step by step:
1. Time taken by A and B to reach Q:
- Speed of A = 3 km/hr
- Speed of B = 4 km/hr
- Distance = 21 km
- Time taken by A = Distance/Speed = 21/3 = 7 hours
- Time taken by B = Distance/Speed = 21/4 = 5.25 hours
2. Meeting point R:
- When B reaches Q, he immediately returns towards P.
- By the time B reaches R, A would have covered a distance of 21 km.
- Since B takes 5.25 hours to reach Q and back, A covers 3 km/hr x 5.25 hrs = 15.75 km by the time they meet at R.
- Remaining distance from R to Q covered by B = 21 - 15.75 = 5.25 km
3. Calculating distance from P to R:
- Total distance from P to Q = 21 km
- Distance covered by A when B reaches Q = 21 km
- Distance covered by B from Q to R = 5.25 km
- Therefore, distance from P to R = 21 - 5.25 = 15.75 km
Hence, the distance from P to R is 15.75 km, which is closest to option B (18 km).

A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters?
  • a)
    1777 m
  • b)
    1822 m
  • c)
    400 m
  • d)
    1400 m
Correct answer is option 'C'. Can you explain this answer?

Ananya Patel answered
Given information:
- Speed of train = 100 kmph
- Speed of motorbike = 64 kmph
- Time taken to overtake = 40 seconds

Calculating relative speed:
- Relative speed = (100 - 64) kmph = 36 kmph
- Convert relative speed to m/s: 36 kmph = 10 m/s

Calculating distance covered in 40 seconds:
- Distance = Speed x Time
- Distance = 10 m/s x 40 s = 400 meters

Length of the train:
- The distance covered includes the length of the train and the motorbike
- Let's assume the length of the train is 'x' meters
- Distance covered by the train = Distance covered by motorbike + Length of the train
- 400 = 64 x (40/3600) + x
- 400 = 7.11 + x
- x = 392.89 meters
Therefore, the length of the train is approximately 400 meters (option 'C').

Two athletes cover the same distance at the rate of 10 and 15 kmph respectively. Find the distance travelled when one takes 15 minutes longer than the other.
  • a)
    8.5 km
  • b)
    750 km
  • c)
    7.5 km
  • d)
    15 km
Correct answer is option 'C'. Can you explain this answer?

Rajeev Kumar answered
The distance travelled is 7.5 km.
Let the time taken by the athlete travelling at 10 kmph be t hours.
The time taken by the athlete travelling at 15 kmph is t -15/60 hours.
The distance travelled by both athletes is the same.
Therefore, 10t = 15(t -15/60)
Solving for t, we get t = 3/4 hours.
The distance travelled by both athletes is 10t = 10 * 3/4 = 7.5 km.

A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
  • a)
    3km
  • b)
    4km
  • c)
    5km
  • d)
    6km
Correct answer is option 'D'. Can you explain this answer?

If a car covers a certain distance at x kmph and an equal distance at y kmph,
the average speed of the whole journey = 2xy/x+y kmph
Hence, average speed = 2*3*2/2+3 = 12/5 km/hr
Total time taken = 5hours
⇒ Distance travelled = 12/5*5 = 12 km
⇒ Distance between his house and office = 12/2 =  6km

A distance is covered at a certain speed in a certain time. If the double of this distance is covered in four times the time, then what is the ratio of the two speeds?
  • a)
    1.5 : 0.7
  • b)
    1 : 1.9
  • c)
    4 : 2
  • d)
    6 : 1
Correct answer is option 'C'. Can you explain this answer?

Upsc Toppers answered
Case I : Distance D Speed S1 Time D/S1
Case II : Distance 2D Speed S2 Time 4(D/S1)
=> Speed for case II = S2 = Distance/Time = 2D/(4D/S1) = S1/22/(4/1) = 1/2
Hence, speed for case I : speed for case II = S1:S= 1:1/2 = 2:1 => Option C is correct.

Find the time taken by two trains, one 180 m long and the other 270 m long, to cross each other, if they are running at speeds of 46 kmph and 54 kmph respectively. Consider both possible cases of motion.
  • a)
    202.5, 16.2 sec
  • b)
    160, 100 sec
  • c)
    108.45, 15.6 sec
  • d)
    204.5, 14.8 sec
Correct answer is option 'A'. Can you explain this answer?

Given data:
- Length of first train (L1) = 180 m
- Length of second train (L2) = 270 m
- Speed of first train (S1) = 46 kmph
- Speed of second train (S2) = 54 kmph

Approach:
- When the two trains are moving in the same direction, the relative speed is the difference between the speeds of the two trains.
- When the two trains are moving in opposite directions, the relative speed is the sum of the speeds of the two trains.

Case a: Trains moving in the same direction
- Relative speed = 54 kmph - 46 kmph = 8 kmph = 8 * 5/18 m/s = 20/9 m/s
- Total distance to be covered = 180 m + 270 m = 450 m
- Time taken = Total distance / Relative speed = 450 / (20/9) = 202.5 seconds

Case b: Trains moving in the opposite direction
- Relative speed = 54 kmph + 46 kmph = 100 kmph = 100 * 5/18 m/s = 250/9 m/s
- Total distance to be covered = 180 m + 270 m = 450 m
- Time taken = Total distance / Relative speed = 450 / (250/9) = 16.2 seconds

Conclusion:
- The time taken for the two trains to cross each other is 202.5 seconds when they are moving in the same direction and 16.2 seconds when they are moving in opposite directions.

If Sita walks at 5 kmph, she misses her train by 10 minutes. If she walks at 7 kmph, she reaches the station 10 minutes early. How much distance does she walk to the station?
  • a)
    5.8 km
  • b)
    35.6 km
  • c)
    10.6 km
  • d)
    92 km
Correct answer is option 'A'. Can you explain this answer?

Rajeev Kumar answered
The distance to the station can be calculated as follows:

Let's denote the distance to the station as "d" (in km), and the time difference between the two cases as "t" (in minutes).

In the first case, Sita walks at 5 km/h and misses the train by 10 minutes. So the time it would take her to get to the train on time is: d/5 (in hours) + 10/60 (in hours) = d/5 + 1/6 (in hours).

In the second case, Sita walks at 7 km/h and arrives 10 minutes early. So the time it takes her to get to the train is: d/7 - 10/60 = d/7 - 1/6 (in hours).

Since these two times should be the same, we can equate them:

d/5 + 1/6 = d/7 - 1/6

Solving this equation for "d" gives:

d = 35/6 km = 5.8 km

So the correct answer is 5.8 km.

A ship is 156 km away from the bank of river. A leak, which admits metric tons of water in  min, but the pumps throughout 15 metric tons in 1 hour. 68 metric tons would sufficient to sink the ship. Find the average rate of sailing so that she may just reach the bank as she begins to sink
  • a)
    15
  • b)
    60
  • c)
    18
  • d)
    10
Correct answer is option 'A'. Can you explain this answer?

In one minute, amount flowing in = 15/39 MT (MT stands for metric tons)
In one minute, amount thrown out = 15/60 = 1/4 MT
Effective rate of filling in one hour = (15/39 - 1/4) MT = 21/56 MT/Min
Time till it just begins to sink = 68 / (21/156)
Speed required = (156/505) = 0.3 km/min = 0.3 x 60 km/hr = 18 km/hr.

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