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 statistics moving averages moving average and variation simple moving average uses of average in day-to-day life introduction to time series introduction to time series and forecasting

Choose the correct answer from the alternatives given.
A man covers a distance of $160$ km at $64 km/hr$ and next $160$ km at $80 km/hr$. What is his average speed for his whole journey of $320$ km?

  1. $73.00km/h$
  2. $71.11km/h$
  3. $70.50km/h$
  4. $72.25km/h$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average speed is total distance divided by total time. Total distance is 160 + 160 = 320 km. Time 1 = 160/64 = 2.5 hours. Time 2 = 160/80 = 2 hours. Total time = 4.5 hours. Average speed = 320 / 4.5 = 71.11 km/h.

Multiple choice vedic methods of multiplication history of mathematics maths

Two trains starts from stations $A$ and $B$ and travel towards each other at speed of $50\ km/hr$ and $60\ km/hr$ respectively. At the time of their meeting, the second train has travelled $120\ km$ more than the first. The distance between $A$ and $B$ is

  1. $990\ km$
  2. $1200\ km$
  3. $1320\ km$
  4. $1440\ km$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Speed of train $A = 50\ kmph$
Speed of train $B = 60\ kmph$
Since, time is constant
$Speed \propto$ Distance covered
$\dfrac {S _{A}}{S _{B}} = \dfrac {D _{A}}{D _{B}}$
$\dfrac {50}{60} = \dfrac {D _{A}}{D _{B}}\Rightarrow \dfrac {5}{6} = \dfrac {D _{A}}{D _{B}}$
Given that train $B$ has travelled $120\ km$ extra.
$6x - 5x = 120$
$x = 120$
The distance between $A$ and $B = 6x + 5x = 11x$
$= 11\times 120 = 1320$
Alternate Method:
Let train $A$ start form station $A$ and $B$ from station $B$.
Let the trains $A$ and $B$ meet after/ hours.
$\therefore$ Distance covered by train $A$ in $t$ hours $= 50t$
Distance covered by train $B$ in $t$ hours $= 60t\ km$.
According to the question,
$60t - 50t = 120$
$\Rightarrow t = \dfrac {120}{10} = 12\ hours$.
$\therefore$ Distance $AB = 50\times 12 + 60 \times 12$
$= 600 + 720 = 1320\ km$.

Multiple choice conversion of time time conversion time measurement of time maths

A boat travels upstream from $B$ to $A$ and down stream from $A$ to $B$ in $3$ hours. If the speed of the boat in still water is $9\ km/hr$ and the speed of the current is $3\ km/ hr$, then the distance (in km) between $A$ and $B$ is

  1. $12$
  2. $8$
  3. $6$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the distance between $A$ and $B$ be $x$ km.
Upstream speed $= 9 - 2 = 6\ kmph$
Downstream speed $= 9 + 3 = 12\ kmph$
$\dfrac {x}{6} + \dfrac {x}{12} = 3\Rightarrow \dfrac {2x + x}{12} = 3\Rightarrow 3x = 36 \Rightarrow x = 12\ km$.

Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

Fighter planes  $X$  and  $Y$  are moving towards a target $ 'O'$ along two perpendicular paths, with equal speeds.  $X$  starts from a point at a distance of  $19\mathrm { km }$  from  $^ { \prime } O ^ { \prime }$  and  $Y$  starts from a point at a distance of   $12\mathrm { km }$  from  $ \mathrm '{ O } '.$  After  $1$  minute, it was found that they were  $13\mathrm { km }$  away from each other. What is the speed at which they are travelling. given that they start simultaneously?

  1. $35 \mathrm { km } / \mathrm { min }$
  2. $28 \mathrm { km } / \mathrm { min }$
  3. $7 \mathrm { km } / \mathrm { min }$
  4. $21 \mathrm { km } / \mathrm { min }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

A car travels 120 km from A to B at 30 km per hour but returns the same distance at 40 km per hour. The average speed for the round trip is closest to: 

  1. 33 km/hr

  2. 34 km/hr

  3. 35 km/hr

  4. 36 km /hr

  5. 37 km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If a car travels a distance d at rate $r _1$  and returns the same distance at rate $r _2$ , then 
Average speed = $\dfrac{total distance}{total time} = \dfrac{2d}{d/r _1+ d/r _2}= \dfrac{2r _1r _2}{r _1+r _2}$; 
$\therefore  x = \dfrac{2.30.40}{70} = \dfrac{240}{7} ~ 34 km/hr$.

Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

Mohit travelled $225$ km in $\displaystyle 4\frac {1}{2}$  hours. Raj travelled the same distance at an average speed that was $10$ km per hour faster than Mohit's average speed. What was the number of hours it took Raj to travel $225$ km ?

  1. $\displaystyle 2\frac {1}{4}$
  2. $\displaystyle 3\frac {1}{4}$
  3. $\displaystyle 3\frac {3}{4}$
  4. $\displaystyle 4\frac {2}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the formula, $Speed=\frac { Distance }{ time } $
Mohit speed=
${ S } _{ m }=\frac { 225 }{ 4.5 } =50km/h$
${ S } _{ r }=50+10=60km/h$
Distance=225km
$Time=\frac { Distance }{ Speed } =\frac { 225 }{ 60 } =3\frac { 3 }{ 4 } hr$
Answer (C) $3\frac { 3 }{ 4 } hr$

Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

Two trains from two places start running in the opposite direction and reach the destination at the midpoint after  $3\dfrac{1}{3}$ hours and $4\dfrac{4}{5}$  hours. If the speed of the first train is $80$ km/hr, then the speed of the 2nd train (in km/hr) is

  1. $64\dfrac{2}{3}$
  2. $66\dfrac{2}{3}$
  3. $55\dfrac{5}{9}$
  4. $75$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\text{Let speed of second train is x km/hr}$

$\text{if both train reaches the destination at the mid point then,}$

$\text{distance left to be covered by A = distance covered by B }.......(i)$
and
$\text{Distance = speed  x time taken}$

from (i)
$x\times\dfrac{24}{5}= 80 \times \dfrac{10}{3}$
$x=55\dfrac{5}{9}$

option c is correct


Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

A boat takes $19$ hours for travelling downstream from point $A$ to point $B$ and coming back to a point $C$ midway between $A$ and $B$. If the velocity of the stream is $4$ $kmph$ and the speed of the boat in still water is $14 \mathrm { kmph}, $ what is the distance between $A$ and $B$?

  1. $160km$
  2. $180km$
  3. $200km$
  4. $220km$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Speed in downstream =  $(14 + 4) km/hr = 18 km/hr ;$

Speed in upstream =  $(14 – 4) km/hr = 10 km/hr. $

Let the distance between A and B be $x$ km. Then,

$\frac{x}{18} + \frac{x}{2}\times\frac{1}{10}$ = 19 

$\Rightarrow \frac{x}{18} + \frac{x}{20}$ = 19 

$\Rightarrow$ $x = 180 km.$

Multiple choice physics study of sound determination of speed of sound by echo problems on sound sound needs a medium

A boy clapped his hands near a cliff and heard the echo after 4s. What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 ms$^{-1}$?

  1. 2 m

  2. 1384 m

  3. 692 m

  4. 2768 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, Speed of sound, $v = 346 ms^{-1}$ Time taken for hearing the echo, $t = 4 s$
Distance travelled by the sound $ = v \times t = 346 m/s \times 4 s = 1384 m$
In 4 s sound has to travel twice the distance between the cliff and the person.
Hence, the distance between the cliff and the person $= 1384 m/2 = 692 m$

Multiple choice physics propagation of sound waves longitudinal vs transverse wave sound and light comparison of speed of sound with speed of light

A worker lives at a distance of 1.32 km from the factory. If the speed of sound in air is 330 m$s^{-1}$, how long will the sound of factory siren take to reach the worker?

  1. 1 s

  2. 2 s

  3. 3 s

  4. 4 s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,


Distance $=1.32\,km=1320\,m$

Speed of sound in air $=330\,m/s$

We have,

$Time=\frac{Distance}{Speed}$

$Time=\frac{1320}{330}=4\,s$

Thus it would take $4\,s$ for the sound of the factory siren to reach the 
worker.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

A man walks a distance of 48 km in a given time. If he walks 2 km/hr faster, he will perform the journey 4 his before. His normal rate of walking is _______.

  1. 3 km/hr

  2. 4 km/hr
    • 6 km/hr or 4 km/hr

  4. 5 km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let us assume that the man walks at a speed of $x$ $kmph$

Then the time taken by the man to complete the journey is $t=\frac{48}{x}$  ($\because time=\frac {distance}{speed}$)
Now in the second case, $t'=t-3$ and $x'=x+2$.
Then, 
$t-4=\frac{48}{x+2}$
$\Rightarrow$ $t-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{48}{x}-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{96}{x(x+2)}=4$
$\Rightarrow$ $x^{2}+2x=24$
$\Rightarrow$ $x^{2}+2x-24=0$
$\Rightarrow$ $x^{2}-4x+6x-24=0$
$\Rightarrow$ $(x-4)(x+6)=0$
Speed cannot be a negative value and hence $x=4kmph$
$\therefore$ Option B is correct.

Multiple choice physics motion and measurement of distances motion around us motion and rest moving things around us

A boy running at an average speed of $4 km hr^{-1}$ reaches school from his home in $\frac{1}{2}hr$. The distance of the school from his home is

  1. $2 km$
  2. $8 km$
  3. $4 km$
  4. $6 km$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:$v =4\frac{Km}{h}$
           

     $t =\dfrac{1}{2}$hour
   as $v=\dfrac{Distance}{t}$
        $Distance=4\times\dfrac{1}{2}$
        $Distance=2Km$