Quantitative Aptitude

Time, Speed and Distance

2,165 Questions
All Core formulas and unit conversion 441 Boats and streams effective speed 396 Relative speed, meetings, and overtaking 375 Average speed 362 Early and late arrival 321 Proportional relationships 202 Stoppage time per hour 35 Train and stationary or moving object crossing 17 Races and dead heats 16

Time, Speed and Distance Questions

Multiple choice

Train A leaves from P to Q. Half an hour later train B leaves for P from Q. Two hours after A had started, the trains meet. A has covered 50 % more distance than train B when they meet. What is the ratio of the speeds of train A to train B?
  1. $9:8$
  2. $8:9$
  3. $5:7$
  4. $8:3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let speed of A be Va and B be Vb. A travels for 2 hours, B travels for 1.5 hours. Distance Da = 2Va, Db = 1.5Vb. Da = 1.5 * Db => 2Va = 1.5 * 1.5Vb => 2Va = 2.25Vb => Va/Vb = 2.25/2 = 9/8.

Multiple choice

Two men undertake to drive a distance of $54$ km. The first performs the journey at $8$ km/hr. The second, starting half an hour later, arrives $15$ minutes sooner. Find the approximate ratio of the speeds of the first to the second person.
  1. $8 : 9$
  2. $8 : 5$
  3. $8 : 6$
  4. $6 : 9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First person time: 54/8 = 6.75 hours. Second person starts 0.5 hours later and arrives 0.25 hours earlier, so their travel time is 6.75 - 0.5 - 0.25 = 6 hours. Speed of second person = 54/6 = 9 km/hr. Ratio of speeds = 8:9.

Multiple choice

Avinash covered 150 km distance in 10 hours. The first part of his journey he covered by car, then he hired a rickshaw. The speed of car and rickshaw is 20 km/hr and 12 km/hr respectively. The ratio of distances covered by car and the rickshaw respectively are

  1. 2 : 3

  2. 4 : 5

  3. 1 : 1

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let d1 be distance by car and d2 by rickshaw. d1 + d2 = 150 and d1/20 + d2/12 = 10. Substituting d2 = 150 - d1: d1/20 + (150 - d1)/12 = 10. Multiply by 60: 3d1 + 5(150 - d1) = 600 => 3d1 + 750 - 5d1 = 600 => 2d1 = 150 => d1 = 75. Then d2 = 75. Ratio is 75:75 = 1:1.

Multiple choice

An ant moves with a uniform speed along a straight line path from P to Q. 60 cm apart, in front of a mirror and takes two minutes to cover this distance. The line segment PQ is inclined to the mirror $M{M}'$ at an angle of $30^{\circ}$. What is the rate at which the distance between the ant and its image changes?
  1. $0.5 cm s^{-1}$
  2. $1.5 cm s^{-1}$
  3. $1.0 cm s^{-1}$
  4. $2.5 cm s^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ant moves at speed v = 60 cm / 120 s = 0.5 cm/s. The angle to the mirror is 30 degrees. The component of velocity perpendicular to the mirror is v * sin(30) = 0.5 * 0.5 = 0.25 cm/s. The image moves at the same speed in the opposite direction, so the relative speed between the ant and its image is 2 * 0.25 = 0.5 cm/s.

Multiple choice

The driver of an express train travelling at a speed of $v_{1}$ sees on the same track at distance $d$ in front of him a goods train travelling in the same direction at a speed $v_{2}$ such that $v_1>v_2$. Immediately he applies brakes to his express train producing retardation $a$ to avoid collision. Then
  1. $a < \dfrac{v^{2}_{1}-v^{2}_{2}}{2d}$
  2. $a < \dfrac{(v_{1}-v_{2})^{2}}{2d}$
  3. $a > \dfrac{(v_{1}-v_{2})^{2}}{2d}$
  4. $a > \dfrac{v^{2}_{1}-v^{2}_{2}}{2d}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To avoid collision, the relative velocity (v1 - v2) must be reduced to zero before the distance d is covered. Using the kinematic equation v_final^2 = v_initial^2 - 2ad, where v_final = 0, v_initial = v1 - v2, we get 0 = (v1 - v2)^2 - 2ad. Thus, a = (v1 - v2)^2 / (2d). To avoid collision, the retardation must be greater than this value.

Multiple choice

Two trains each travelling at a speed of $20$ kmph approach each other on the same straight track. A bird that can fly at a speed of $40$ kmph flies off from one train when they are $40$ km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train and so forth. Before the trains crash, the total distance travelled by the bird is
  1. $20 km$
  2. $40 km$
  3. $80 km$
  4. $60 km$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The trains are 40 km apart and each travels at 20 kmph. Their relative speed is 20 + 20 = 40 kmph. The time until they crash is 40 km / 40 kmph = 1 hour. The bird flies at 40 kmph for that entire hour, so it travels 40 kmph * 1 hour = 40 km.

Multiple choice

A car travels $\dfrac{1}{3}^{rd}$ distance on a straight road with a velocity of 10 km/hr, next $\dfrac{1}{3}^{rd}$ with velocity 20 km/hr and the last $\dfrac{1}{3}^{rd}$ with velocity 60 km/hr. What is the average speed of the car in the whole journey?

  1. 4 km/hr

  2. 6 km/hr

  3. 12 km/hr

  4. 18 km/hr
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average speed = Total Distance / Total Time. Let total distance be 3D. Time = D/10 + D/20 + D/60 = (6D+3D+D)/60 = 10D/60 = D/6. Average speed = 3D / (D/6) = 18 km/hr.

Multiple choice

A $1500 kg$ car is initially moving with speed $30 {m}/{s}$. Suddenly, a rock falls onto the road ahead. The maximum coefficient of friction between the car tires and the roads is $\mu = 0.7$. Approximately how much distance will the car need to stop to avoid hitting the rock?
  1. $22 m$
  2. $46 m$
  3. $66 m$
  4. $87 m$
  5. $131 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the work-energy theorem, the kinetic energy 1/2 * m * v^2 is dissipated by friction force f * d = mu * m * g * d. Thus, 1/2 * v^2 = mu * g * d. d = v^2 / (2 * mu * g) = 30^2 / (2 * 0.7 * 9.8) = 900 / 13.72 = 65.59 m, which is approximately 66 m.