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All questions of Speed, Distance and Time for RRB NTPC/ASM/CA/TA 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?

Raghavendra Sharma answered
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
Clear all your RRB NTPC/ASM/CA/TA doubts with EduRev

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?

Impact Learning 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?

EduRev CAT 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 train of 300 m is travelling with the speed of 45 km/h when it passes point A completely. At the same time, a motorbike starts from point A with the speed of 70 km/h. When it exactly reaches the middle point of the train, the train increases its speed to 60 km/h and motorbike reduces its speed to 65 km/h. How much distance will the motorbike travel while passing the train completely?
  • a)
    2.52 km
  • b)
    2.37 km
  • c)
    2 km
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Shalini Bajaj answered
Speed of train while passing point
A = 70 x (5/18) m/s = VI
Speed of bike initially = 70 x (5/18) m/s = V2
Time taken by the bike to reach at the mid-point of the train = 1 5 0 /(V 2 - V I)
Again find out the new speeds of train and bike, and calculate the time taken by the bike to cover the rest 150 m distance relative to the train.

Two riders on the horseback with a gun and a bullet proof shield were moving towards each other at a constant speed of 20 km/h and 5 km/h respectively. When they were 100 km apart, they started firing bullets at each other at the speed of 10 km/h. When a bullet of rider 1 hits the shield of rider 2, rider 2 fires a bullet and the process continues vice versa. Neglecting the time lag at the instant when the bullet hits the shield and the rider fires the shot, find the total distance covered by all the bullets shot by both the riders.
  • a)
    50 km
  • b)
    40 km
  • c)
    25 km
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Yash Khanna answered
This question has actually got nothing to do with bullets initially. We can see that it takes them 4 hours to reach each other. And this is the same time for which bullets will cover some distance.
So, the total distance covered by the bullet = 4 x 1 0 = 40 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?

Pallabi Deshpande answered
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.

There are two swimmers A and B who start swimming towards each other from opposite banks of the lake. They meet at a point 900 ft from one shore for the first time. They cross each other, touch the opposite bank and return. They meet each other again at 300 ft from the other shore. What is the width of the lake?
  • a)
    2400 ft
  • b)
    1800 ft
  • c)
    2700 ft
  • d)
    3600 ft
Correct answer is option 'A'. Can you explain this answer?

Aditi Kumar answered
Let us assume that the width of the lake = x. So, when one of the runners A covers 900 m, the other one B is covering (x - 900) m. To meet next time, A will be covering (x - 900 + 300) m whereas B will be covering (900 + X-300) m.
Now, 900/(x - 900) = (x - 900 + 300)/(x + 900 - 300)
Now use options to find the answer.

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

Distance between Lucknow and Patna is 300 km. Mayank leaves at a speed of x km/h from Lucknow towards Patna. After three hours Sharat leaves at the speed of (x + 10) km/h from Lucknow towards Patna. If x and the number of hours taken to meet after Sharat starts are integers, how much distance can Mayank cover before they meet?
  • a)
    174 km
  • b)
    60 km
  • c)
    150 km
  • d)
    180 km
Correct answer is option 'B'. Can you explain this answer?

Deepika Mukherjee answered
One of the ways of solving this question is going through equations. But after a certain stages we will be required to start assuming the values because all the data are not given.
Another way of doing this problem is: Start working by assuming some values. Let us assume the speed of Mayank =10 km/h. In three hours he has covered 30 km. Now Sharat starts with a speed of 20 km/h. He will take 3 hours to meet Mayank. Till that time, the total distance covered by Mayank = 60 km.

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?

Aspire Academy answered
Eight hours for a 600 km journey, when 120 km is done by train and 480 km by car.
It takes 20 minutes more if 200 km is done by train and 400 km by car.
Formula used:
Speed = Distance/Time
Calculation:
Let the speed of the train be x km/h
And the speed of the car be y km/h
⇒ 120/x + 480/y = 8
⇒ 120(1/x + 4/y) = 8
⇒ 1/x + 4/y = 1/15     ...i)
In the second condition
⇒ Total time = 8 + 20/60 = 25/3 hr
∴  200/x + 400/y = 25/3
⇒ 200(1/x + 2/y) = 25/3
⇒ 1/x + 2/y = 1/24     ...ii)
After solving equation (i) and (ii)
(By substracting equation 2 from equation 1)
⇒ x = 60 km/h
⇒ y = 80 km/h
Ratio of the speed of train and car is
⇒ 60 : 80
⇒ 3 : 4
∴ The ratio of the speed of train and car is 3 : 4.

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.

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.

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 merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an overall collection and the value is linked to the number of items that are already on the market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs. 6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?
    • a)
      9
    • b)
      16
    • c)
      27
    • d)
      36
    Correct answer is option 'C'. Can you explain this answer?

    Rajeev Kumar answered
    Let us assume he buys n goods.
    Total CP = 20n
    Total SP = 2 + 4 + 6 + 8 ….n terms
    Total SP should be at least 40% more than total CP
    2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n
    2 (1 + 2 + 3 + ….n terms) ≥ 28n
    n(n + 1) ≥ 28n
    n2 + n ≥ 28n
    n2 - 27n ≥ 0
    n ≥ 27
    The question is " If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?"
    He should sell a minimum of 27 goods.
    Hence, the answer is 27.
    Choice C is the correct answer.

    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?

    Arindam Mukherjee answered
    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').

    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?

    Raghavendra Sharma answered
    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 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?

    Pallabi Deshpande answered
    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:S2 = 1:1/2 = 2:1 => Option C is correct.

    Read the passage given below and solve the questions based on it There arc two cities Nagpur and Pune along a straight line 120 km apart. A and B start at the same time from Nagpur and Pune respectively w ith the speed of 40 km/h and 60 km/h respectively. They travel towards each other, and after they meet for the first time, they reverse directions and also interchange their speeds. After reaching their respective starting points, they reverse their directions and start proceeding towards each other again.
    Q.
    How many hours from the beginning do they meet for the second time?
    • a)
      2 h
    • b)
      2*1/5 h
    • c)
      2*1/3 h
    • d)
      3*3/5 h
    Correct answer is option 'A'. Can you explain this answer?

    Sagar Sharma answered
    Given Information:
    - There are two cities, Nagpur and Pune, that are 120 km apart along a straight line.
    - A and B start at the same time from Nagpur and Pune, respectively.
    - A's speed is 40 km/h and B's speed is 60 km/h.
    - They travel towards each other and after they meet for the first time, they reverse directions and interchange their speeds.
    - After reaching their respective starting points, they reverse their directions and start proceeding towards each other again.

    To Find:
    The number of hours from the beginning when they meet for the second time.

    Solution:
    Let's calculate the time taken by A and B to meet for the first time.

    - The total distance between Nagpur and Pune is 120 km.
    - The relative speed of A and B when they are traveling towards each other is the sum of their individual speeds, which is 40 km/h + 60 km/h = 100 km/h.
    - The time taken to cover a distance of 120 km at a speed of 100 km/h is 120 km / 100 km/h = 1.2 hours.

    After they meet for the first time:
    - A and B reverse their directions and interchange their speeds.
    - Now A's speed is 60 km/h and B's speed is 40 km/h.
    - The total distance between Nagpur and Pune remains the same, which is 120 km.
    - The relative speed of A and B when they are traveling towards each other after the first meeting is the sum of their individual speeds, which is 60 km/h + 40 km/h = 100 km/h.
    - The time taken to cover a distance of 120 km at a speed of 100 km/h is 120 km / 100 km/h = 1.2 hours.

    So, the total time taken for A and B to meet for the second time is 1.2 hours + 1.2 hours = 2.4 hours.

    However, the question asks for the number of hours from the beginning when they meet for the second time. Since they started at the same time, the number of hours from the beginning when they meet for the second time is half of the total time taken, which is 2.4 hours / 2 = 1.2 hours.

    Answer:
    Therefore, they meet for the second time after 1.2 hours from the beginning. Option A is the correct answer.

    Read the passage given below and solve the questions based on it There are 8 days in a week from Sunday io Saturday and another day called Fund ay on planet North. There are 36 h in a day and each hour has 90 min while each minute has 60 s.
    Q.
    Find the approximate angle between the hands of a clock on North when the time is 16.50 am?
    • a)
      189
    • b)
      131
    • c)
      320
    • d)
      165
    Correct answer is option 'B'. Can you explain this answer?

    Nayanika Basu answered
    Method to Solve :

    36hrs a day mane the clk will b of 18hrs.

    18hrs covers 360 deg

    Thus 1hr covers 20  deg

    12 hrs covers 12*20=240 deg

    1 hr in 90min

    Thus in 90 min the hour hand goes 20 deg

    For 1min hr hand goes 2/9deg

    For 40 min hour  goes (2/9)*40 deg=80/9 deg=9 deg (approx)

    thus hr hand covers total of 240+9=249 deg

    now min hand covers=40*4=160 deg

    (90min=360deg

    1min=4deg)

    Thus angle between them 249-160 = 89 deg

    Ramesh and Somesh are competing in a 100 m race. Initially, Ramesh runs at twice the speed of Somesh for the first fifty m. After the 50 m mark, Ramesh runs at l/4th his initial speed while Somesh continues to run at his original speed. If Somesh catches up with Ramesh at a distance of ‘N ’ m from the finish line, then N is equal to
    • a)
      35
    • b)
      10
    • c)
      45
    • d)
      None of these
    Correct answer is option 'D'. Can you explain this answer?

    Pallabi Deshpande answered
    This question gives us the freedom to assume any value of speeds of Ramesh and Somesh. 
    Let us assume the initial speed of Somesh = 20 m/s, then the initial speed of Ramesh = 40 m/s.
    Till 50 m they are running with this speed only. 

    Time taken by Ramesh in covering 50m = 1.25sec. In the same time Somesh is covering 25m. 
    After this stage, the speed of Somesh is 20m/s,  
    whereas speed ofRasmesh = 10 m/s. 

    Now relative speed = 10m/s and distance = 25m. 
    At 75m from the starting, both of them will be meeting.

    Refer to the data below and answer the questions that follow.
    There is a race between Sagar and Sapna. Both of them Delhi. Sagar started on a bike with a speed of 40 km/h and Sapna has started in a car with a speed of 60 km/h from Mumbai to Delhi After 31/2 h of the journey there was a snag in the car She tried to repair the car but in vain After half an hour she got a lift for another 500 km in a truck, which was travelling with a speed of 45 km/h. From there Delhi was at a distance of 200 km on road, instead, Sapna took a shorter route which was only 100 km away from Delhi She started running at the speed of 30 km/h on the shorter route to reach Delhi.
    Q. If there was no snag in the car, by how much distance Sapna would have defeated Sagar?
    • a)
      252 km
    • b)
      264 km
    • c)
      303 km
    • d)
      321 km
    Correct answer is option 'C'. Can you explain this answer?

    Anaya Patel answered
    Total distance from Mumbai to Delhi = 910 km Time taken by Sapna to cover this distance, had there been no snag = 910/60
    Since their speeds are in the ratio 3:2, Manoj would have covered [910 - (910 x 40)/60] km less than Sapna.

    Read the passage below and solve the questions based on it* A number o f runners, numbered 1, 2. 3 , , N and so on, start simultaneously at the same point on a circular track and keep on running continuously in the same direction, around the track. They run in such a way that the speed of the runner numbered N (N>1) isN times that of the runner numbered 1 .
    Q.
    If there are exactly six runners, then at how many distinct points on the track is the runner numbered 1 overtaken by any of the other five runners?
    • a)
      15
    • b)
      11
    • c)
      9
    • d)
      10
    Correct answer is option 'D'. Can you explain this answer?

    Maulik Rane answered
    Assume that the track length is 1000 m.
    Now, runner 1 and runner 2 will meet at one point, i.e., the starting point.
    Runner 1 and runner 3 will meet at two points, at 500 m and at the starting point.
    Runner 1 and runner 4 will meet at three points, at 333.33 m, at 666.66 m and at the starting point. Runner 1 and runner 5 will meet at four points, at 250 m, 500 m, 750 m and at starting point.
    Runner 1 and runner 6 will meet at five points, at 200 m, at 400 m, at 600 m, at 800 m and at the starting point.
    These are 10 distinct points.

    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.

    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?

    Lohit Matani answered
    Case I: Motion in same direction
    =>Relative speed = 54 — 46 = 8 km/hr.
    Distance to be covered = 180 + 270 = 450 m.
    =>Time = 0.450/8 = 0.056 hrs = 202.5 sec.
     
    Case II: Motion in opposite direction.
    => Relative Speed = 54 + 46=100 km/hr.
    Distance to be covered = 180 + 270= 450 m.
    =>Time = 0.450/100 = 0.00045 hrs = 16.2 sec.

    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?

    Master Training Institute answered
    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.

    Read the passage given below and solve the questions based on it Two cars A and B start simultaneously from two different cuies P and Q and head for the alies 0 and P respectively, As soon as car A reaches the city Q, it turns and starts for city P and as soon as il reaches P again, it leaves for city Q and so on. Similarly car B, travels from Q to P, from P to Q and so on. The speeds of the car A and B are in the ratio o f 3 :4.
    Q.
    If car Atravelled 1860 m, how many times did the cars A and B meet? (Take the distance PQ as calculated in the previous question.)
    • a)
      14
    • b)
      15
    • c)
      16
    • d)
      Cannot be determined
    Correct answer is option 'C'. Can you explain this answer?

    Preethi Nair answered
    If A covers 1860 m, then in the same time B covers 2480 m (ratio of speed 3:4). So, total distance covered = 4340 m = 140 m + 30 x 140 m = 31 x 140 m, this is the distance covered for 16th meeting.

    A train approaches a tunnel PQ which is 16 m long. Two rabbits A and B are standing at points which are 12 m and 8 m inside the tunnel with respect to the entrance P. When the train is x m away from P, A starts running towards P and B towards Q. Difference between the ratios of the speed of A to that of the train and the ratio of the speed of B to that of train is 1/8. How much can the distance x be, if both of them get caught at the ends of the tunnel?
    • a)
      30 m
    • b)
      16 m
    • c)
      48 m
    • d)
      None of these
    Correct answer is option 'C'. Can you explain this answer?

    Gowri Chakraborty answered
    tunnel pq=16m, 
    a(s)= speed of rabbit a , similarly b(s) for rabbit b;
    t(s)= speed of train.
    now a and b are sitting at 12m  and 8m away from the enterance from which train is coming, they will move to opposite ends.a towards p and b towards q and are caught by train at those ends. so we can say
    when train moves x distance, a will move 12m and when train move x+16, b will move 8m.
    now a(s)/t(s)-b(s)/t(s)=1/8
    or time is same , use distance only in the speed formula
    12/x-8/x+6=1/8; put the value of x from the options or make quadratic equation.
    x=48 m

    Chapter doubts & questions for Speed, Distance and Time - Mathematics for RRB NTPC / ASM 2025 is part of RRB NTPC/ASM/CA/TA exam preparation. The chapters have been prepared according to the RRB NTPC/ASM/CA/TA exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for RRB NTPC/ASM/CA/TA 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

    Chapter doubts & questions of Speed, Distance and Time - Mathematics for RRB NTPC / ASM in English & Hindi are available as part of RRB NTPC/ASM/CA/TA exam. Download more important topics, notes, lectures and mock test series for RRB NTPC/ASM/CA/TA Exam by signing up for free.

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