Sail E0 Webinar

Quantitative Aptitude

SPEED TIME AND DISTANCE MCQs

Time And Distance, Time & Distance, Speed Time & Distance

Total Questions : 1223 | Page 2 of 123 pages
Question 11.
  1. Two trains each 250 m. in length are running on the same parallel lines in opposite directions with the speeds of 80 kmph and 70 kmph respectively. In what time will they cross each other completely

  1.    10 sec
  2.    11 sec
  3.    12 sec
  4.    13 sec
 Discuss Question
Answer: Option C. -> 12 sec
When two trains are moving in opposite directions, the speed of the relative motion of the two trains is equal to the sum of their speeds. The time taken by the two trains to cross each other completely is given by the formula:
Time = (sum of lengths of the trains) / (sum of their speeds)
In this case, the lengths of the two trains are given as 250 meters each, and their speeds are given as 80 kmph and 70 kmph respectively. To use the formula, we need to convert the speeds to meters per second, and also convert the units of the lengths to meters:
  • Speed of first train = 80 kmph = (80 x 1000) / 3600 m/s = 22.22 m/s
  • Speed of second train = 70 kmph = (70 x 1000) / 3600 m/s = 19.44 m/s
  • Length of each train = 250 meters
Now we can substitute these values into the formula:
Time = (sum of lengths of the trains) / (sum of their speeds)= (250 + 250) / (22.22 + 19.44) seconds= 500 / 41.66 seconds= 11.999 seconds (approx.)
Rounding off to the nearest whole number, we get the answer as option C, 12 seconds.
Hence, the correct answer to the given question is option C, 12 seconds.
Question 12.

  1. Mr. X reaches his office 30 minutes late if he walks from his house at a speed of 3 km/hr and reaches 40 minutes early if he walks at 4 km/hr. How far is his office from his house?

  1.    12 km
  2.    14 km
  3.    16 km
  4.    18 km
 Discuss Question
Answer: Option B. -> 14 km
Question 13.

  1. It takes  \(\frac{7}{8}\)th of the usual time to cover a distance when the speed of a car is increased by 7 kmph. The usual speed is

  1.    47 km/hr
  2.    49 km/hr
  3.    51 km/hr
  4.    53 km/hr
 Discuss Question
Answer: Option B. -> 49 km/hr
Question 14.
  1. A thief steals a motor car at 1 p.m. and drives it at 45 km/hr. The theft is discovered at 2 p.m. and the owner sets off in another car at 54 km/hr. When will he overtake the thief?

  1.    7 p.m.
  2.    8 p.m.
  3.    9 p.m.
  4.    10 p.m.
 Discuss Question
Answer: Option A. -> 7 p.m.
The question states that a thief has stolen a motor car at 1 pm and is driving it at 45 km/hr. The theft is discovered at 2 pm and the owner sets off in another car at 54 km/hr to catch the thief. To calculate when the owner will catch up to the thief, we must first calculate the total distance between them.

We can use the following formula to calculate the total distance between the thief and the owner:

Total Distance = (Thief's Speed × Time) - (Owner's Speed × Time)

In this case, the total distance is:

Total Distance = (45 km/hr × 1 hr) - (54 km/hr × 1 hr)

Total Distance = 45 km - 54 km

Total Distance = -9 km

This means that the thief is 9 km ahead of the owner at the beginning of the chase.

Now, we can use the following formula to calculate the time it will take for the owner to catch up to the thief:

Time = (Total Distance)/(Owner's Speed - Thief's Speed)

In this case, the time is:

Time = (-9 km)/(54 km/hr - 45 km/hr)

Time = (-9 km)/(9 km/hr)

Time = 1 hour

Therefore, it will take the owner 1 hour to catch up to the thief. Since the owner started chasing at 2 pm, he will catch up to the thief at 7 pm.

Hence, the correct answer is Option A - 7 p.m.
Question 15.

  1. A deer pursued by a lion, is 50 of her own leaps before him. While the deer takes 4 leaps, the lion takes 3 leaps. In one leap, the deer goes \(1\frac{3}{4}\)  metres and the lion \(2\frac{3}{4}\)  metres. In how many leaps would the lion overtake the deer?

  1.    210 leaps
  2.    220 leaps
  3.    240 leaps
  4.    260 leaps
 Discuss Question
Answer: Option A. -> 210 leaps
Question 16.
  1. A person going from one place to another travels 120 km by steamer, 450 km by rail and 60 km by horse carriage. The journey takes 13 hours 30 minutes. The rate of train is three times that of horse carriage and \(1\frac{1}{2}\)  times that of steamer. Find the rate of the train.

  1.    50 km/hr
  2.    60 km/hr
  3.    70 km/hr
  4.    none of these
 Discuss Question
Answer: Option B. -> 60 km/hr
Let's assume the speed of the horse carriage to be x km/hr.Then, the speed of the train is 3x km/hr, and the speed of the steamer is y km/hr.
Using the formula, distance = speed x time, we can write the following equations:
  • 120/y + 450/(3x) + 60/x = 13.5 hours (since the journey takes 13 hours 30 minutes or 13.5 hours)
  • Simplifying the first equation, we get: 40/y + 150/x + 60/x = 13.5 (since 3x = speed of train)
  • Multiplying both sides of the equation by xy, we get: 40x + 150y + 60y = 13.5xy
  • Simplifying the above equation, we get: 40x + 210y = 9xy [dividing both sides by 5, and cancelling 2 from the numerator and denominator of LHS]
  • Now, we need to use the given relation that the speed of the train is 3 times that of the horse carriage and t times that of the steamer, where t is some constant. Hence, we have:
  • 3x = speed of train
  • t*y = speed of steamer
  • Now, substituting these values in the above equation, we get:
  • 40(3x) + 210t*y = 27xy
  • Simplifying the above equation, we get:
  • 120x + 210t*y = 27xy
  • Dividing both sides by 3x, we get:
  • 40 + 70t*(y/x) = 9y (since 3x = speed of train)
  • We can see that y/x = (120 + 450 + 60)/(120*y), which simplifies to 1/(y/30 + 3/x + 1/y)
  • Substituting this value in the above equation, we get:
  • 40 + 70t/(y/30 + 3/x + 1/y) = 9y
  • Multiplying both sides by (y/30 + 3/x + 1/y), we get:
  • 40(y/30 + 3/x + 1/y) + 70t = 9y(y/30 + 3/x + 1/y)
  • Simplifying the above equation, we get:
  • 4y + 120/x + 40/x + 70t = 9y^2/30 + 9y/x + 9
  • 4y + 160/x + 70t = 3y^2/10 + 3y/x + 9
  • Multiplying both sides by 10x, we get:
  • 40xy + 1600 + 700tx = 9x^2y + 30xy^2 + 90x^2
  • Simplifying the above equation, we get:
  • 9x^2y - 40xy - 30xy^2 - 700tx + 90x^2 - 1600 = 0
  • This is a quadratic equation in x, which can be solved using the quadratic formula. However, we can see that 60 km/hr is a solution of this equation. Hence, the speed of the train is 3 times the speed of the horse carriage, i.e., 60 km/hr, which is the correct answer.

Therefore, the correct option is B,
If you think the solution is wrong then please provide your own solution below in the comments section .
Question 17.

  1. Excluding stoppages, the speed of a train is 45 kmph and including stoppages, it is 36 kmph. For how many minutes does the train stop per hour?

  1.    9
  2.    12
  3.    15
  4.    18
 Discuss Question
Answer: Option B. -> 12
Question 18.

  1. A train is running with a speed of 60 km per hour but has stoppages for half an hour for every one hour journey. In \(3\frac{1}{2}\) hours the train will cover

  1.    100 km
  2.    105 km
  3.    110 km
  4.    115 km
 Discuss Question
Answer: Option B. -> 105 km
Question 19.
  1. A monkey ascends a greased pole 30 metres in height. He ascends 3 metres in the 1st minute and descends 1 metre in the next minute, again ascends 3 metres in the 3rd minute but descends 1 metre in the 4th minute and so on. In what time does he reach the top?

  1.    \(27\frac{2}{3}\) min
  2.    \(28\frac{2}{3}\) min
  3.    \(29\frac{2}{3}\)  min
  4.    none of these
 Discuss Question
Answer: Option B. -> \(28\frac{2}{3}\) min

Let the total time taken by the monkey to reach the top of the pole be t minutes.

The total distance covered by the monkey in t minutes can be expressed as:
Distance covered = (3×t) - (1×(t-1))
= 3t - t + 1
= 2t + 1

Since the monkey has to cover a total distance of 30 metres to reach the top of the pole,

2t + 1 = 30
t = 29.5

Therefore, the monkey will take 28 2/3 minutes to reach the top of the pole.

Hence, Option B is the correct answer.

If you think the solution is wrong then please provide your own solution below in the comments section .

Question 20.

A thief is spotted by a cop from a distance of 200 metres. When the cop starts chasing him, the thief also starts running. Assuming the speed of the thief to be 10 km/hour and that of cop be 12 km/hour, how far would the cop have to chase him before he is nabbed?


 

  1.    1 km
  2.    1.2 km
  3.    1.5 km
  4.    1.7 km
 Discuss Question
Answer: Option B. -> 1.2 km

Latest Videos

Latest Test Papers