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

A swimmer can swim in still water with speed $v$ and the river is flowing with speed $v/ 2$. What is the ratio of the time taken to swimming across the river in the shortest time to that of swimming across the river over the shortest distance?
  1. $\frac{{\sqrt 3 }}{2}$
  2. $\frac{1}{2}$
  3. $2$
  4. $\sqrt 3 v$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Shortest time (t1) occurs when swimming perpendicular to the bank: t1 = width / v. Shortest distance (t2) occurs when the resultant velocity is perpendicular to the bank: t2 = width / sqrt(v^2 - (v/2)^2) = width / (v * sqrt(3)/2). The ratio t1/t2 = (width/v) / (width * 2 / (v * sqrt(3))) = sqrt(3)/2.

Multiple choice

A car, 'A' is moving towards south with a speed of 20 m/s. Another car 'B' is moving towards east with a speed of 15 m/s. At a certain instant, the car 'A' is due south of the car 'B' and is at distance of 50 m from car B. The shortest distance between the car 'A' & 'B'.

  1. 30 m

  2. 10 m

  3. 40 m

  4. 35 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assuming the cars are moving towards each other along perpendicular paths, their relative distance squared at time t is given by D^2 = (15t)^2 + (50 - 20t)^2 = 625t^2 - 2000t + 2500. Minimizing this quadratic expression by setting its derivative to zero gives t = 1.6 seconds. Substituting t = 1.6 back into the distance formula yields a minimum distance of sqrt(900) = 30 meters.

Multiple choice

A car travelling at a speed of $30\ km/h$ is brought to rest in a distance of $8\ m$ by applying brakes. If the same car is moving at a speed of $60\ km/h$ then it can be brought to rest with same brakes in
  1. $64\ m$
  2. $32\ m$
  3. $16\ m$
  4. $4\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done by brakes equals change in kinetic energy. (1/2)mv^2 = F*d. Since F is constant, d is proportional to v^2. If speed doubles (30 to 60), distance increases by 2^2 = 4 times. 8 m * 4 = 32 m.

Multiple choice

If the body covers one-third distance at speed $v_{1}$, next one third at speed $v_{2}$ and last one third at speed $v_{3}$, then average speed will be
  1. $\dfrac {v_{1}v_{2} + v_{2}v_{3} + v_{3}v_{1}}{v_{1} + v_{2} + v_{3}}$
  2. $\dfrac {v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2}v_{3} + v_{3}v_{1}}$
  3. $\dfrac {v_{1} + v_{2} + v_{3}}{3}$
  4. $\dfrac {3v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2}v_{3} + v_{3}v_{1}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average speed is total distance divided by total time. Let total distance be 3d. Time taken for each segment is d/v1, d/v2, and d/v3. Total time = d(1/v1 + 1/v2 + 1/v3) = d(v2v3 + v1v3 + v1v2)/(v1v2v3). Average speed = 3d / [d(v1v2 + v2v3 + v3v1)/(v1v2v3)] = 3v1v2v3 / (v1v2 + v2v3 + v3v1).

Multiple choice

A tourist being chased by an energy bear is running in a straight line towards his car at a speed of $4.0 \ m/s$. The car is a distance $d$ away. The bear is $26 \ m$ behind the tourist and running at $6.0 \ m/s$. The tourist reaches the car safely the maximum possible value for $d$ is:-
  1. $25 \ m$
  2. $26 \ m$
  3. $52 \ m$
  4. $50 \ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Tourist speed = 4 m/s, Bear speed = 6 m/s. Relative speed = 2 m/s. The bear covers the 26m gap in 26/2 = 13 seconds. In 13 seconds, the tourist covers 4 * 13 = 52 meters. Thus, the car must be at most 52 meters away.

Multiple choice

A car moves for half of its time at $80Km/hr$ and for rest of time $40 Km/hr$. If total distance covered is $60Km,$ the average sped of the car is:
  1. $120Km/hr$
  2. $100Km/hr$
  3. $80Km/hr$
  4. $60Km/hr$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average speed = Total distance / Total time. Let total time be 2T. Time at 80 km/h is T, time at 40 km/h is T. Distance = 80T + 40T = 120T. Given 120T = 60, so T = 0.5 hours. Total time = 1 hour. Average speed = 60 km / 1 hour = 60 km/h.

Multiple choice

A car covers a distance of $2 \ km$ in $2.5$ minutes. If it covers half of the distance with speed $40 \ km/hr$, the rest distance it shall cover with a speed of:
  1. $56 \ km/hr$
  2. $60 \ km/hr$
  3. $48 \ km/hr$
  4. $50 \ km/hr$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total distance = 2 km. Time = 2.5 min = 2.5/60 hr = 1/24 hr. Half distance = 1 km. Time for first half = 1 km / 40 km/hr = 1/40 hr. Remaining time = 1/24 - 1/40 = (5-3)/120 = 2/120 = 1/60 hr. Speed for second half = 1 km / (1/60 hr) = 60 km/hr.

Multiple choice

A car has to cover the distance $60\ km$. If half of the total time it travels with speed $80\ km/h$ and in rest half time its speed becomes $40\ km/h$, the average speed of car will be:
  1. $60\ km/h$
  2. $80\ km/h$
  3. $120\ km/h$
  4. $180\ km/h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Average speed is total distance divided by total time. Let total time be T. Distance covered in first half time is 80 * (T/2) = 40T. Distance in second half is 40 * (T/2) = 20T. Total distance = 60T = 60 km, so T = 1 hour. Average speed = 60 km / 1 hour = 60 km/h.