CAT Time Distance Work Questions & Solutions

A sample of real CAT Time Distance Work past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 99 Time Distance Work questions in all — sign up free to practise them timed.

Q1.CAT 2025
Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is
- 76800
- 112000
- 96000
- 86400
Show solution
According to question, Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours.

Given, in the first part time required is 30 minutes.

Also, the time taken to cover these four parts follow arithmetic progression.

So, let us say the times be 30 minutes, (30+d) minutes, (30+2d) minutes and (30+3d) minutes

So, $30+\left(30+d\right)+\left(30+2d\right)+\left(30+3d\right)=180$

or, $6d=180-120$

or, $6d=60$

or, $d=\dfrac{60}{6}=10$ minutes.

So, the time required in these four parts are 30 minutes, 40 minutes, 50 minutes and 60 minutes respectively.

Now, in the first part, speed is 960 metres per minute.

Speed in the four parts is also in arithmetic progression.

Let's say the speeds be $(960+x)$ , $(960+2x)$ , $(960+3x)$ metres per minute

So, total distance covered = $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60$ metres

So, $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60=224000$

or, $960\left(30+40+50+60\right)+40x+\left(2x\right)\left(50\right)+\left(3x\right)\left(60\right)=224000$

or, $172800+320x=224000$

or, $320x=224000-172800=51200$

or, $x=\dfrac{51200}{320}=160$

So, the speed in the fourth part is $960+3x=960+3\times\ 160=1440$ metres per minute

Time in the fourth part is 60 minutes

So, distance covered in the fourth part = $1440\times\ 60=86400$ metres

So, option D is the correct answer.
Q2.CAT 2023
Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of $x$ km till then, where $x$ is a whole number and is palindromic, i.e., $x$ remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was
- 110
- 90
- 100
- 80
Show solution
Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.
Q3.CAT 2022
Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be
- 4
- 12
- 8
- 6
Show solution

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 - 8 = 16

In triangle OFS, $\cos60^0\ =\ \frac{OF}{OS}$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 - 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.
Q4.CAT 2020
Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual total time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is
- 29
- 26
- 28
- 27
Show solution
Let distance = d

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

=> d = 28km

The actual time taken to travel 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 = $\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\$ km/hr
Q5.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
Q6.CAT 2018
Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is
- 2 : 7
- 2 : 9
- 1 : 2
- 1 : 4
Show solution
Car 3 meets car 1 at Q, which is 200 km from A.
Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.
As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.
Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.
As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1
And speed of car 2 : speed of car 3 = 1 : 2
So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer.
Q7.CAT 2008
Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
- 6 : 15 am
- 6 : 30 am
- 6 :45 am
- 7 : 00 am
- 7 : 15 am
Show solution

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.
Hence, $\angle\ ABC=60^{\circ\ }$ , $\angle\ CAB=30^{\circ\ }$ , which implies $\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$
Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.
We know that $\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$
Similarly, $\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$

So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around $\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.
Q8.CAT 2004
A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is $\pi$ r during the first 30 seconds, $\pi$ r/2 during next one minute, $\pi$ r/4 during next 2 minutes, $\pi$ r/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?
- 4
- 8
- 16
- 32
Show solution
Let radius be 1 units and p = 3.14 or $\pi$ . So circumference is $2*\pi$ .

According to given condition distance covered in first 1/2 mins = $\pi$ /2 km, distance covered in next 1 min = $\pi$ /2 km, distance covered in next 2 mins = $\pi/2$ km and finally distance covered in next 4 minutes = $\pi/2$ km.

Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes.

Now time taken to cover $\pi/2$ is in GP.

For the second round the time taken is = 8+16+32+64 = 120

Ratio = 120/7.5 = 16
Q9.CAT 2001
At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?
- $\frac{7}{3}$
- $\frac{4}{3}$
- $\frac{5}{3}$
- $\frac{8}{3}$
Show solution
$12/(R - S) = T$
$12/(R + S) = T - 6$
$12/(2R - S) = t$
$12/(2R + S) = t - 1$
=> $12/(R - S) - 12/(R + S) = 6$ and $12/(2R - S) - 12/(2R + S) = 1$
=> $12R + 12S - 12R + 12S = 6R^2 - 6S^2$ and $24R + 12S - 24R + 12S = 4R^2 - S^2$
=> $24S = 6R^2 - 6S^2 and 24S = 4R^2 - S^2$
=> $6R^2 - 6S^2 = 4R^2 - S^2$
=> $2R^2 = 5S^2$
=> $24S = 10S^2 - S^2 = 9S^2$
=> $S = 24/9 = 8/3$
Q10.CAT 1999
Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 am and travels at 50km per hour towards Baroda situated 100 kms away. At 7:00 am Howrah - Ahmedabad express leaves Baroda towards Ahmedabad and travels at 40 km per hour. At 7:30 Mr. Shah, the traffic controller at Baroda realises that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?
- 15 minutes
- 20 minutes
- 25 minutes
- 30 minutes
Show solution
The distance between Ahmedabad and Baroda is 100 Km
Navjivan express starts at 6:30 am at 50 Km/hr and Howrah expresses starts at 7:00 am at 40 Km/hr.
Distance covered by Navjivan express in 30 minutes (by 7 am) is 25 Km/hr.

So, at 7 am, the distance between the two trains is 75 Kms and they are travelling towards each other a relative speed of 50+40=90 Km/hr.

So, time taken them to meet is 75/90*60 = 50 minutes.

As, Mr. Shah realizes the problem after thirty minutes, time left to avoid collision is 50-30 = 20 minutes
Q11.CAT 1998
I started climbing up the hill at 6 a.m. and reached the top of the temple at 6 p.m. Next day I started coming down at 6 a.m. and reached the foothill at 6 p.m. I walked on the same road. The road is so short that only one person can walk on it. Although I varied my pace on my way, I never stopped on my way. Then which of the following must be true?
- My average speed downhill was greater than that of uphill
- At noon, I was at the same spot on both the days.
- There must be a point where I reached at the same time on both the days.
- There cannot be a spot where I reached at the same time on both the days.
Show solution
The total time taken for both the journeys is the same. So, a) is incorrect. If he goes with the same speed both ways, he would've been at the same point at noon on both days. So, d) is incorrect. If he does not travel with the same speed in both the directions, he need not be at the same spot at noon. So, option b) is incorrect. Option c) is the correct answer.
Passage
A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr.
Q12.CAT 1997
At what time did X catch the thief?
- 3.30 p.m.
- 3 p.m.
- 3.15 p.m.
- None of these
Show solution
Time taken to catch the thief = $\frac{d}{v}$ (Where d is relative distance achieved in 15 min. = 15km. and v is relative velocity of poiliceman i.e. = 65-60 = 5 kmph)
So time taken = 3hr.
Hence he will catch thief at 3:15 pm

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CAT Time Distance Work previous year questions with solutions — AthenaPrep

CAT Time Distance Work Questions & Solutions

Q1.CAT 2025

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is

76800
112000
96000
86400

Show solution

According to question, Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours.

Given, in the first part time required is 30 minutes.

Also, the time taken to cover these four parts follow arithmetic progression.

So, let us say the times be 30 minutes, (30+d) minutes, (30+2d) minutes and (30+3d) minutes

So, $30+\left(30+d\right)+\left(30+2d\right)+\left(30+3d\right)=180$

or, $6d=180-120$

or, $6d=60$

or, $d=\dfrac{60}{6}=10$ minutes.

So, the time required in these four parts are 30 minutes, 40 minutes, 50 minutes and 60 minutes respectively.

Now, in the first part, speed is 960 metres per minute.

Speed in the four parts is also in arithmetic progression.

Let's say the speeds be $(960+x)$ , $(960+2x)$ , $(960+3x)$ metres per minute

So, total distance covered = $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60$ metres

So, $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60=224000$

or, $960\left(30+40+50+60\right)+40x+\left(2x\right)\left(50\right)+\left(3x\right)\left(60\right)=224000$

or, $172800+320x=224000$

or, $320x=224000-172800=51200$

or, $x=\dfrac{51200}{320}=160$

So, the speed in the fourth part is $960+3x=960+3\times\ 160=1440$ metres per minute

Time in the fourth part is 60 minutes

So, distance covered in the fourth part = $1440\times\ 60=86400$ metres

So, option D is the correct answer.

Q2.CAT 2023

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of $x$ km till then, where $x$ is a whole number and is palindromic, i.e., $x$ remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was

Show solution

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.

Q3.CAT 2022

Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

Show solution

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 - 8 = 16

In triangle OFS, $\cos60^0\ =\ \frac{OF}{OS}$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 - 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.

Q4.CAT 2020

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

Show solution

Let distance = d

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

=> d = 28km

The actual time taken to travel 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 = $\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\$ km/hr

Q5.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

Q6.CAT 2018

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

2 : 7
2 : 9
1 : 2
1 : 4

Show solution

Car 3 meets car 1 at Q, which is 200 km from A.
Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.
As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.
Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.
As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1
And speed of car 2 : speed of car 3 = 1 : 2
So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer.

Q7.CAT 2008

Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to

6 : 15 am
6 : 30 am
6 :45 am
7 : 00 am
7 : 15 am

Show solution

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.

Hence, $\angle\ ABC=60^{\circ\ }$ , $\angle\ CAB=30^{\circ\ }$ , which implies $\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$

Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.

We know that $\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$

Similarly, $\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$

So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around $\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.

Q8.CAT 2004

A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is $\pi$ r during the first 30 seconds, $\pi$ r/2 during next one minute, $\pi$ r/4 during next 2 minutes, $\pi$ r/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?

Show solution

Let radius be 1 units and p = 3.14 or $\pi$ . So circumference is $2*\pi$ .

According to given condition distance covered in first 1/2 mins = $\pi$ /2 km, distance covered in next 1 min = $\pi$ /2 km, distance covered in next 2 mins = $\pi/2$ km and finally distance covered in next 4 minutes = $\pi/2$ km.

Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes.

Now time taken to cover $\pi/2$ is in GP.

For the second round the time taken is = 8+16+32+64 = 120

Ratio = 120/7.5 = 16

Q9.CAT 2001

At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?

$\frac{7}{3}$
$\frac{4}{3}$
$\frac{5}{3}$
$\frac{8}{3}$

Show solution

$12/(R - S) = T$
$12/(R + S) = T - 6$
$12/(2R - S) = t$
$12/(2R + S) = t - 1$
=> $12/(R - S) - 12/(R + S) = 6$ and $12/(2R - S) - 12/(2R + S) = 1$
=> $12R + 12S - 12R + 12S = 6R^2 - 6S^2$ and $24R + 12S - 24R + 12S = 4R^2 - S^2$
=> $24S = 6R^2 - 6S^2 and 24S = 4R^2 - S^2$
=> $6R^2 - 6S^2 = 4R^2 - S^2$
=> $2R^2 = 5S^2$
=> $24S = 10S^2 - S^2 = 9S^2$
=> $S = 24/9 = 8/3$

Q10.CAT 1999

Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 am and travels at 50km per hour towards Baroda situated 100 kms away. At 7:00 am Howrah - Ahmedabad express leaves Baroda towards Ahmedabad and travels at 40 km per hour. At 7:30 Mr. Shah, the traffic controller at Baroda realises that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?

15 minutes
20 minutes
25 minutes
30 minutes

Show solution

The distance between Ahmedabad and Baroda is 100 Km
Navjivan express starts at 6:30 am at 50 Km/hr and Howrah expresses starts at 7:00 am at 40 Km/hr.
Distance covered by Navjivan express in 30 minutes (by 7 am) is 25 Km/hr.

So, at 7 am, the distance between the two trains is 75 Kms and they are travelling towards each other a relative speed of 50+40=90 Km/hr.

So, time taken them to meet is 75/90*60 = 50 minutes.

As, Mr. Shah realizes the problem after thirty minutes, time left to avoid collision is 50-30 = 20 minutes

Q11.CAT 1998

I started climbing up the hill at 6 a.m. and reached the top of the temple at 6 p.m. Next day I started coming down at 6 a.m. and reached the foothill at 6 p.m. I walked on the same road. The road is so short that only one person can walk on it. Although I varied my pace on my way, I never stopped on my way. Then which of the following must be true?

My average speed downhill was greater than that of uphill
At noon, I was at the same spot on both the days.
There must be a point where I reached at the same time on both the days.
There cannot be a spot where I reached at the same time on both the days.

Show solution

The total time taken for both the journeys is the same. So, a) is incorrect. If he goes with the same speed both ways, he would've been at the same point at noon on both days. So, d) is incorrect. If he does not travel with the same speed in both the directions, he need not be at the same spot at noon. So, option b) is incorrect. Option c) is the correct answer.

Passage

A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr.

Q12.CAT 1997

At what time did X catch the thief?

3.30 p.m.
3 p.m.
3.15 p.m.
None of these

Show solution

Time taken to catch the thief = $\frac{d}{v}$ (Where d is relative distance achieved in 15 min. = 15km. and v is relative velocity of poiliceman i.e. = 65-60 = 5 kmph)
So time taken = 3hr.
Hence he will catch thief at 3:15 pm

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