# CAT Time Speed and Distance Questions PDF [Most Important]

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Time Speed and Distance is one of the most important topics in the CAT Quants section. If you’re weak in Time Speed and Distance questions for CAT, make sure you learn all the basic concepts of solving the questions and also all the important formulas of TSD. Here, you can learn all the important concepts in CAT Time Speed and Distance. You can check out these CAT Time Speed and Distance questions from the CAT Previous year’s papers.Â

This post will look at important Time Speed & Distance questions in the CAT quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT TSD Questions PDF along with the detailed solutions (and video solutions) below, which is completely Free.

Question 1:Â Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for â‚¹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

Solution:

Now Anil Paints in 12 Days
Barun paints in 16 Days
Now together Arun , Barun and Chandu painted in 6 Days
Now let total work be W
Now each worked for 6 days
So Anil’s work = 0.5W
Barun’s work =Â $\frac{6W}{16}=\frac{3W}{8}$
Therefore Charu’s work =Â $\frac{W}{2}-\frac{3W}{8}=\ \frac{W}{8}$
Therefore proportion of charu =$\frac{24000}{8}=\ 3,000$

Question 2:Â Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

Solution:

Considering the distance travelled by Mira in one minute = M,

The distance traveled by Amal in one minute = A.

Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)

C is the circumference of the circle.

Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :

Hence 45*(A-M) = 3C. (2)

Multiplying (1)*15 we have :

45A + 45M = 15C.

45A – 45M = 3C.

Adding the two we have A = $\frac{18C}{90}$

Subtracting the two M = $\frac{12C}{90}$

Since Mira travels $\frac{12C}{90}$ in one minute, in one hour she travels :$\frac{12C}{90}\cdot60\ =\ 8C$

Hence a total of 8 rounds.

Alternatively,

Let the length of track be L
and velocity of Mira be a and Amal be b
Now when they meet after 45 minutes Amal completes 3 more rounds than Mira
so we can say they met for the 3rd time moving in the same direction
so we can say they met for the first time after 15 minutes
So we know Time to meet = Relative distance /Relative velocity
so we get $\frac{15}{60}=\frac{L}{a-b}$ Â  Â  (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $\frac{3}{60}=\frac{L}{a+b}$ Â Â  (2)
Dividing (1) and (2)
we get $\frac{\left(a+b\right)}{\left(a-b\right)}=5$
or 4a =6b
or a = 3b/2
Now substituting in (1)
we get :
$\frac{L}{b}\times\ 2=\ \frac{15}{60}$
so $\frac{L}{b}\ =\frac{1}{8}$
So we can say 1 round is covered in $\frac{1}{8}$ hours
so in 1-hour total rounds covered = 8.

Question 3:Â One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

a)Â 11.5

b)Â 10

c)Â 12.5

d)Â 12

Solution:

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.

Question 4:Â Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is

a)Â 144

b)Â 140

c)Â 264

d)Â 120

Solution:

Let A fill the tank at x liters/hour and B drain it at y liters/hour
Now as per Condition 1 :
We get Volume filled till 10pm = 8x-7y Â  Â  Â  (1) .
Here A operates for 8 hours and B operates for 7 hours .
As per condition 2
We get Volume filled till 6pm = 4x-2y Â  Â  Â Â  (2)
Here A operates for 4 hours and B operates for 2 hours .
Now equating (1) and (2)
we get 8x-7y =4x-2y
so we get 4x =5y
y =4x/5
So volume of tank =Â $8x-7\times\ \frac{4x}{5}=\frac{12x}{5}$
So time taken by A alone to fill the tank = $\frac{\frac{12x}{5}}{x}=\frac{12}{5}hrs\$
= 144 minutes

Question 5:Â Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is

a)Â 3:2

b)Â 5:3

c)Â 2:3

d)Â 2:1

Solution:

Considering the length of train A = La, length of train B = Lb.

The speed of train A be 5*x, speed of train B be 3*x.

From the information provided :

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.

Hence (8*x)*(46) = Lb.

In the information provided :

It took another 69 seconds for the rear ends of the trains to cross each other.

In the next 69 seconds

The train B traveled a distance equivalent to the length of train A in this 69 seconds.

Hence (8*x)*(69) = La.

La/Lb = 69/46 = 3/2 = 3 : 2

Question 6:Â Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

Solution:

Let the total work be 48 units. Let Amar do ‘m’ work, Akbar do ‘k’ work, and Anthony do ‘n’ units of work in a month.

Amar and Akbar complete the project in 12 months.Â Hence, in a month they doÂ $\frac{48}{12}$=4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we getÂ $m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$.

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would takeÂ $\frac{48}{1.5}=32$months.

Question 7:Â Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. IfÂ the speed of the other train is 6 km/hr less than the faster one, its length, in m, is

a)Â 184

b)Â 192

c)Â 190

d)Â 180

Solution:

Speed of the faster train =Â $\frac{160}{12}=\frac{40}{3}\$ m/s

Speed of the slower train =Â $\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$ m/s

Sum of speeds (when the trainsÂ travel towards each other) =Â $\frac{40}{3}+\frac{35}{3}=25$ m/s

Let the slower train be $x$ metres long; then:Â $\frac{160+x}{25}=14$

On solving,Â $x=190\ m$

Question 8:Â Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A.Â Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2Â at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram willÂ make before he meets Rahim again for the first time is

a)Â 5

b)Â 3

c)Â 2

d)Â 4

Solution:

To complete one round Ram takes 100m/15kmph and Rahim takes 20m/5kmph

They meet for the first time after L.C.M of (100m/15kmph , 20m/5kmph) = 100m/5kmph=20m/kmph.

Distance traveled by Ram =20m/kmph * 15kmph =300m.

So, he must have ran 300/100=3 rounds.

Note:

CAT gave both 2 and 3 as correct answers because of the word ‘before‘.

Question 9:Â A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other. Ram and Rahim reach their destination in one minute and four minutes, respectively. if they start at the same time, then the ratio of Ram’s speed to Rahim’s speed is

a)Â $\frac{1}{2}$

b)Â $\sqrt{2}$

c)Â $2$

d)Â $2\sqrt{2}$

Solution:

Let the speed of Ram be v(r) and the speed of Rahim be v(h) respectively. Let them meet after time “t” from the beginning.

Hence Ram will cover v(r)(t) during that time and Rahim will cover v(h)t respectively.

Now after meeting Ram reaches his destination in 1 min i.e. Ram covered v(h)t in 1 minute or v(r)(1)= v(h)(t)

Similarly Rahim reaches his destination in 4 min i.e. Rahim covered v(r)t in 4 minutes or v(h)(4)= v(r)(t)

Dividing both the equations we getÂ $\frac{v\left(r\right)}{4v\left(h\right)}=\frac{v\left(h\right)}{v\left(r\right)}\ or\ \frac{v\left(r\right)}{v\left(h\right)}=2$ Hence the ratio is 2.

Question 10:Â In a car race, car A beats car B by 45 km. car B beats car C by 50 km. and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

a)Â 475

b)Â 450

c)Â 500

d)Â 550

Solution:

Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).

$\frac{v\left(a\right)}{v\left(b\right)}=\frac{m}{m-45}$ …..(1)where ‘”m” is the entire distance of the race track.

MoreoverÂ $\frac{v\left(b\right)}{v\left(c\right)}=\frac{m}{m-50}$…….(2)

and finallyÂ $\frac{v\left(a\right)}{v\left(c\right)}=\frac{m}{m-90}$……(3)

Multiplying (1) and (2) we get (3).Â $\frac{m}{m-90}=\frac{m}{m-45}\left(\frac{m}{m-50}\right)$

Solving we get m=450 which is the length of the entire race track

Question 11:Â John takes twice as much time as Jack to finish a job. Jack and Jim together takeÂ one-thirds of the time to finish the job than John takes working alone. Moreover, inÂ order to finish the job, John takes three days more than that taken by three of themÂ working together. In how many days will Jim finish the job working alone?

Solution:

Let Jack take “t” days to complete the work, then John will take “2t” days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) .

Now let Jim take “m” days to complete the work. According to question,Â $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$ Hence Jim takes “2t” time to complete the work.

Now let the three of them complete the work in “p” days. Hence John takes “p+3” days to complete the work.

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work

Question 12:Â The distance from B to C is thrice that from A to B. Two trains travel from A to C via B.Â The speed of train 2 is double that of train 1 while traveling from A to B and theirÂ speeds are interchanged while traveling from B to C. The ratio of the time taken byÂ train 1 to that taken by train 2 in travelling from A to C is

a)Â 5:7

b)Â 4:1

c)Â 1:4

d)Â 7:5

Solution:

Let the distance from A to B be “x”, then the distance from B to C will be 3x. Now the speed of Train 2 is double of Train 1. Let the speed of Train 1 be “v”, then the speed of Train 2 will be “2v” while travelling from AÂ to B.

Time taken by Train 1 = (x/v)

Time taken by Train 2 = (x/2v)

Now from B to C distance is “3x” and the speed of Train 2 is (v) and the speed of Train 1 is (2v).

Time taken by Train 1 = 3x/2v

Time taken by Train 2 = 3x/v

Total time taken by Train 1 = x/v(1+(3/2)) = (5/2)(x/v)

Total time taken by Train 2 = x/v(3+(1/2))= (7/2)(x/v)

Ratio of time taken =Â $\frac{5}{\frac{2}{\frac{7}{2}}}=\frac{5}{7}$

Question 13:Â Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the sameÂ point at the same time, and going in the clockwise direction. If they run at speeds of 15Â km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have runÂ when Anil and Sunil meet again for the first time at the starting point?

a)Â 4.8

b)Â 4.6

c)Â 5.2

d)Â 4.2

Solution:

Anil and Sunil will meet at a first point after LCM (Â $\frac{3}{15},\frac{3}{10}$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

Question 14:Â A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, headingÂ towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towardsÂ A at a speed of 20 km/hr. The trains meet each other at

a)Â 11 : 45 am

b)Â 11 : 20 am

c)Â 11 : 00 am

d)Â 10 : 45 am

Solution:

The distance travelled by A between 9:00 Am and 10:30 Am is 3/2*40 =60 km.

Now they are separated by 30 km

Let the time taken to meet =t

Distance travelled by A in time t +Â Distance travelled by B in time t = 30

40t + 20t =30 => t=1/2 hour

Hence they meet at 11:00 AM

Question 15:Â Vimla starts for office every day at 9 am and reaches exactly on time if she drives atÂ her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. OneÂ day, she covers two-thirds of her distance to office in one-thirds of her usual total time toÂ reach office, and then stops for 8 minutes. The speed, in km/hr, at which she shouldÂ drive the remaining distance to reach office exactly on time is

a)Â 29

b)Â 26

c)Â 28

d)Â 27

Solution:

Let distance = d

Given,Â $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

=> d = 28km

The actual timeÂ  taken to travelÂ 28km = 28/40 = 7/10 hours = 42 min.

Given time taken to travel 58/3 km = 1/3 *42 = 14 min.

Then a break of 8 min.

To reach on time, he should cover remaining 28/3 km in 20 min => Speed =Â $\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\$ km/hr

Question 16:Â Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

a)Â 87

b)Â 82

c)Â 78

d)Â 75

Solution:

Let the length of the train be $l kms$ and speed be $s kmph$. Base on the two scenarios presented, we obtain:

$\frac{l}{s-2}=\frac{90}{3600}$….(i) andÂ $\frac{l}{s-4}=\frac{100}{3600}$…(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $s$ asÂ $22 kmph$. Substituting this value in (i), we haveÂ $l=\frac{90}{3600}\times\ 20\ kms$ {keeping it in km and hours for convenience}

Since we need to findÂ $\frac{l}{s}$, let this be equal to $x$. Then,Â $x\ =\ 90\times\frac{20}{22}\ =81.81\ \approx\ 82\ \sec onds\$

Hence, Option B is the correct choice.

Question 17:Â A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

a)Â 50

b)Â 58

c)Â 67

d)Â 61

Solution:

Let the total distance be ‘D’ km and the speed of the train be ‘s’ kmph. The time taken to cover D at speed d is ‘t’ hours. Based on the information: on equating the distance, we getÂ $s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$
On solving we acquire the value ofÂ $\ t\ =\frac{1}{4}$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed ‘s’ {distance traveled in terms of sÂ =Â $\ \frac{s}{12}$}. Remaining distance in terms of ‘s’ =Â $\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $15 – 9 = 6 mins$ or {1/10 hours}. Thus, let ‘x’ kmph be the new value of speed. Based on the above, we getÂ $\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$

Since the increase in speed needs to be calculated:Â $\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$ increase.

Hence, Option C is the correct answer.

Question 18:Â Leaving home at the same time, Amal reaches the office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9.10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?

a)Â 13

b)Â 12

c)Â 14

d)Â 11

Solution:

The difference in the time take to traverse the same distance $’d’$Â at two different speeds is 35 minutes. Equating this:Â $\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$

On solving, we obtain $d = 10 kms$. Let $x kmph$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$.Hence, Option B is the correct answer.

Question 19:Â A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is

a)Â 100

b)Â 90

c)Â 80

d)Â 70

Solution:

Let the speed of Car 2 be ‘x’ kmph and the time taken by the two carsÂ to meet be ‘t’ hours.

In ‘t’ hours, Car 1 travels $\left(60\ \times\ t\right)\ km$ while Car 2 travels $\left(x\ \times\ t\right)\ km$

It is given that the time taken by Car 1 to travel $\left(x\ \times\ t\right)\ km$ is 45 minutes or (3/4) hours. $\therefore\ \frac{\left(x\ \times\ t\right)}{60}\ =\ \frac{3}{4}\$ or $t=\frac{180}{4x}$….(i)

Similarly, the time taken by Car 2 to travel $\left(60\ \times\ t\right)\ km$ is 20 minutes or (1/3) hours. $\therefore\ \frac{\left(60\times\ t\right)}{x}=\frac{1}{3}$ or $\therefore\ t=\frac{x}{180}$….(ii)

Equating the values in (i) and (ii), and solving for x:

$\therefore\ \frac{180}{4x}=\frac{x}{180}\ \ \longrightarrow\ \ \ x\ =90\ kmph$

Hence, Option B is the correct answer.

Question 20:Â One can use three different transports which move at 10, 20, and 30 kmph, respectively to reach from A to B. Amal took each mode of transport for $\frac{1}{3}^{rd}$ of his total journey time, while Bimal took each mode of transport for $\frac{1}{3}^{rd}$ of the total distance. The percentage by which Bimalâ€™s travel time exceeds Amalâ€™s travel time is nearest to

a)Â 22

b)Â 20

c)Â 19

d)Â 21

Solution:

Assume the total distance between A and B as d andÂ time taken by Amal = t

Since Amal travelledÂ $\frac{1}{3}^{rd}$ of his total journey time in different speeds
d =Â $\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$

$\text{Total time taken by Bimal} =Â \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$

$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$

Hence, the ratio ofÂ time taken by Bimal to time taken by Amal = $\frac{\frac{11t}{9}}{t}=\frac{11}{9}$
Therefore, BimalÂ will exceed Amal’s time byÂ $\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$