CAT Time Speed Distance Questions & Solutions

A sample of real CAT Time Speed Distance past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 30 Time Speed Distance questions in all — sign up free to practise them timed.

Q1.CAT 2023
The minor angle between the hours hand and minutes hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8.48 am when this angle increases by 50% is
- $\frac{36}{11}$
- 2
- 4
- $\frac{24}{11}$
Show solution
The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $\dfrac{11}{2}m=12$ => m = $\dfrac{24}{11}$
Q2.CAT 2023
Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is
Answer: 972
Show solution
Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $t^2=6\times\ 24$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.
Q3.CAT 2022
Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is
- 6
- 15
- 10
- 12
Show solution

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$

It is given,

$\dfrac{D}{a}=10$ and $\dfrac{x}{b}=9$

$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$

$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$

$10x^2=\ \ 9D^2-9Dx$

$10x^2+9Dx-9D^2=\ 0$

Solving, we get $x=\dfrac{3D}{5}$

$\dfrac{x}{b}=9$

$\dfrac{3D}{b\times5}=9$

$\dfrac{D}{b}=15$

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B
Q4.CAT 2022
Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance traveled, in km, by the slower car when it meets the other car while traveling towards each other, is
- 100
- 90
- 120
- 150
Show solution
Both the cars take 1.5 hrs to meet when they travel towards each other.

It is given, speed of slower car is 60 km/hr

Therefore, distance covered by slower car before they meet = 60*1.5 = 90 km

The answer is option B.
Q5.CAT 2021
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. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is
- 184
- 192
- 190
- 180
Show solution
Speed of the faster train = $\frac{160}{12}=\frac{40}{3}\$ m/s

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

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

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

On solving, $x=190\ m$
Q6.CAT 2021
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
Answer: 8
Show 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 $\frac{15}{60}=\frac{L}{a-b}$ (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $\frac{3}{60}=\frac{L}{a+b}$ (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.
Q7.CAT 2020
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
- 87
- 82
- 78
- 75
Show 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) and $\frac{l}{s-4}=\frac{100}{3600}$ ...(ii)

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

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

Hence, Option B is the correct choice.
Q8.CAT 2020
A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at
- 11 : 45 am
- 11 : 20 am
- 11 : 00 am
- 10 : 45 am
Show 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 + Distance travelled by B in time t = 30

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

Hence they meet at 11:00 AM
Q9.CAT 2020
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
- 475
- 450
- 500
- 550
Show 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.

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

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

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

Solving we get m=450 which is the length of the entire race track
Q10.CAT 2020
Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2 at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is
- 5
- 3
- 2
- 4
Show 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'.
Q11.CAT 2019
John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?
Answer: 48
Show solution
Speed of John = 6kmph

Speed of Mary = 7.5 kmph

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$

= 48 sec
Q12.CAT 2019
The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was
- $18 \pi$
- $14 \pi$
- $16 \pi$
- $12 \pi$
Show solution
Distance covered by A in 1 revolution = 2 $\pi\$ *30 = 60 $\pi\$

Distance covered by B in 1 revolution = 2 $\pi\$ *40 = 80 $\pi\$

Now, (5000+n)60 $\pi\$ = 80 $\pi\$ n

=> 15000= 4n-3n =>n=15000

Then distance travelled by B = 15000*80 $\pi\$ cm = 12 $\pi\$ km

Hence, the speed = $\ \frac{\ 12\pi\times\ 60\ }{45}$ = 16 $\pi\$

18+ more Time Speed Distance questions inside

Get the full CAT Time Speed Distance set with timed practice, bookmarks and analysis — free to start.

Start practising free

CAT Time Speed Distance previous year questions with solutions — AthenaPrep

CAT Time Speed Distance Questions & Solutions

Q1.CAT 2023

The minor angle between the hours hand and minutes hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8.48 am when this angle increases by 50% is

$\frac{36}{11}$
2
4
$\frac{24}{11}$

Show solution

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $\dfrac{11}{2}m=12$ => m = $\dfrac{24}{11}$

Q2.CAT 2023

Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is

Answer: 972

Show solution

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $t^2=6\times\ 24$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

Q3.CAT 2022

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is

Show solution

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$

It is given,

$\dfrac{D}{a}=10$ and $\dfrac{x}{b}=9$

$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$

$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$

$10x^2=\ \ 9D^2-9Dx$

$10x^2+9Dx-9D^2=\ 0$

Solving, we get $x=\dfrac{3D}{5}$

$\dfrac{x}{b}=9$

$\dfrac{3D}{b\times5}=9$

$\dfrac{D}{b}=15$

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

Q4.CAT 2022

Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance traveled, in km, by the slower car when it meets the other car while traveling towards each other, is

Show solution

Both the cars take 1.5 hrs to meet when they travel towards each other.

It is given, speed of slower car is 60 km/hr

Therefore, distance covered by slower car before they meet = 60*1.5 = 90 km

The answer is option B.

Q5.CAT 2021

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. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is

Show solution

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

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

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

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

On solving, $x=190\ m$

Q6.CAT 2021

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

Answer: 8

Show 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 $\frac{15}{60}=\frac{L}{a-b}$ (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $\frac{3}{60}=\frac{L}{a+b}$ (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.

Q7.CAT 2020

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

Show 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) and $\frac{l}{s-4}=\frac{100}{3600}$ ...(ii)

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

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

Hence, Option B is the correct choice.

Q8.CAT 2020

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

11 : 45 am
11 : 20 am
11 : 00 am
10 : 45 am

Show 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 + Distance travelled by B in time t = 30

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

Hence they meet at 11:00 AM

Q9.CAT 2020

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

Show 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.

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

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

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

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

Q10.CAT 2020

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

Show 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'.

Q11.CAT 2019

John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

Answer: 48

Show solution

Speed of John = 6kmph

Speed of Mary = 7.5 kmph

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$

= 48 sec

Q12.CAT 2019

The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was

$18 \pi$
$14 \pi$
$16 \pi$
$12 \pi$

Show solution

Distance covered by A in 1 revolution = 2 $\pi\$ *30 = 60 $\pi\$

Distance covered by B in 1 revolution = 2 $\pi\$ *40 = 80 $\pi\$

Now, (5000+n)60 $\pi\$ = 80 $\pi\$ n

=> 15000= 4n-3n =>n=15000

Then distance travelled by B = 15000*80 $\pi\$ cm = 12 $\pi\$ km

Hence, the speed = $\ \frac{\ 12\pi\times\ 60\ }{45}$ = 16 $\pi\$

18+ more Time Speed Distance questions inside

Get the full CAT Time Speed Distance set with timed practice, bookmarks and analysis — free to start.

Start practising free

CAT Time Speed Distance previous year questions with solutions — AthenaPrep