0
3369

# 20 Most Important CAT Questions On Time Speed And Distance With Video Solutions PDF (Easy Tips and Tricks to Solve CAT Questions)

Time speed and distance is one of the most commonly appearing concepts in the CAT exam under arithmetic questions. This interesting topic deals with important metrics of a moving object, such as relative speed, average speed, total time and distance etc. The questions from this topic could be a little complex. Aspirants are advised to be well-focused while preparing this concept. It is recommended to practice as many problems as possible from this topic.

In order to help you, we are providing you with some important and must-practice CAT questions from the time speed distance topic, which were taken from the CAT previous years’ papers and will provide a detailed video solution for each question. After practising these questions, you will become a master in this topic and can be able to solve any question from this topic. You can also download these important questions with a video solutions PDF for free by clicking on the link below. The solutions for the below questions are solved and explained by all-time CAT topper and IIM Alumni.

## Quick Tips to Solve Time Speed And Distance Questions in CAT

Time speed and distance is the most important topic for all Competitive examinations. The questions from this topic vary from easy to difficult. The following are the most important formulas and some easy tips and tricks that help you to answer the questions in an easy, fast and accurate way. You can also download these formulas and question-solving tips in a PDF format. Download the free time speed and distance formula PDF.

• Distance = Speed*Time
• Speed = Distance/Time
• Time = Distance/Speed
• While converting the Speed in m/s to km/hr, multiply it by 3.6. It is because 1 m/s = 3.6 km/hr.
• If the ratio of the speeds of A and B is a : b, then
The ratio of the times taken to cover the same distance is 1/a : 1/b or b : a.
The ratio of distance travelled in equal time intervals is a : b.
• Average speed when time is same Sa= S1 +S2 /2
• Average speed when distances are same Sa= 2(S1S2)/ (S1+S2)

## Must-practice Time Speed And Distance Questions for CAT Exam

Question 1:Â A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train’s departure. The distance (in km) from his home to the railway station is

Solution:

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr.

Let d be the distance

Hence,

$\frac{d}{12} – \frac{d}{15} = \frac{1}{3}$

$\frac{d}{60} = \frac{1}{3}$

d = 20 Km.

Question 2:Â A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

a)Â $\sqrt{6}:\sqrt{2}$

b)Â $\sqrt{7}:2$

c)Â $2\sqrt{5}:3$

d)Â 3:2

Solution:

Let the speed of the river be $x$ and the speed of the boat be $u$. Let $d$ be the one way distance and $t$ be the initial time taken.

Given,

$t = \frac{d}{u – x} + \frac{d}{u + x}$ … i

Also,

$\frac{t}{4} = \frac{d}{2u – x} + \frac{d}{2u + x}$

$t = \frac{4d}{2u – x} + \frac{4d}{2u + x}$ … ii

Equating both i and ii,

$\dfrac{d}{u – x}$+ $\dfrac{d}{u + x}$ =Â  $\dfrac{4d}{2u – x} + \dfrac{4d}{2u + x}$

$\dfrac{2u}{u^2 – x^2} = \dfrac{16u}{4u^2 – x^2}$

$4u^2 – x^2 = 8u^2 – 8x^2$

$\frac{u^2}{x^2} = \frac{7}{4}$

$\frac{u}{x} = \frac{\sqrt{7}}{2}$

Question 3:Â Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?

a)Â 1 hour

b)Â 50 minutes

c)Â 1/2 hour

d)Â 55 minutes

Solution:

Since we know that Neera’s husband drives at a uniform speed to and from his residence.
If he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM.
So, Neera must have walked for 55 minutes from 5PM.

Question 4:Â A man travels from A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be

a)Â 3 hr

b)Â 6 hr

c)Â 2 hr

d)Â 4 hr

Solution:

Total time taken to reach at D:

$\frac{12}{x} + x + \frac{12}{2x} + 2x + \frac{12}{4x} = 16$

Or $3x^2 – 16x + 21 = 0$

From the options we can see that only, $x$ = 3hr satisfies the equation. Thus, A is the right choice.

Question 5:Â Three wheels can complete 60, 36 and 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

a)Â $\frac{5}{2}$ s

b)Â $\frac{5}{3}$ s

c)Â $5$ s

d)Â $7.5$ s

Solution:

The first wheel completes a revolution in $\frac{60}{60}=1$ second
The second wheel completes a revolution in $\frac{60}{36}=1\frac{2}{3}$ second
The third wheel completes a revolution in $\frac{60}{24}=2\frac{1}{2}$ second

The three wheels touch the ground simultaneously at time which are multiples of the above three times.
Hence, the required number is $LCM(1,\frac{5}{3},\frac{5}{2}) = 5$ seconds.

So, the correct option is option (c)

Question 6:Â The speed of a railway engine is 42 Km per hour when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 Km per hour when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is:

a)Â 49

b)Â 48

c)Â 46

d)Â 47

Solution:

The function of the speed of the train = 42 – k$\sqrt{n}$ where n is the number of compartments and k is a constant.

42 – k$\sqrt{9}$ = 24

=> 3k = 18 => k = 6

=> Function of speed =Â 42 – 6$\sqrt{n}$

Speed is 0 whenÂ 42 – 6$\sqrt{n}$ = 0

=>Â 42 = 6$\sqrt{n}$

=> n = 49

=> So, with a positive speed, the train can carry 48 compartments.

Question 7:Â 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?

a)Â 15 minutes

b)Â 20 minutes

c)Â 25 minutes

d)Â 30 minutes

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)Â 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

Question 8:Â A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is

a)Â 25, 130

b)Â 30,150

c)Â 20, 90

d)Â None of these

Solution:

Time difference, when second time car’s engine failed at a distance of 30 km., is of 9 min.
Hence putting this in equation:
$\frac{12}{\frac{4v}{5}} – \frac{12}{v} = \frac{9}{60}$ hr. (Because difference of time is considered with extra travelling of 12 km. in second case)
We will get $v (velocity) = 20$ km/hr.
Now for distance $\frac{d-18}{16} + \frac{18}{20} – \frac{d}{20} = \frac{45}{60}$ hr. (As car is 45 min. late after engine’s faliure in first case)
So $d$ = 78 km.
Hence none of these will be our answer.

Question 9:Â Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?

a)Â 12 hours

b)Â 10 hours

c)Â 8 hours

d)Â 6 hour

Solution:

As machine B’s efficiency is twice as of A’s, Hence, it will complete its work in 30 hours.
And C’s efficiency is putting A and B together i.e. = 20 hours $( (\frac{1}{60} + \frac{1}{30})^{-1})$
Now if all three work together, then it will be completed in x (say) days.

$\frac{1}{x} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60}$

or x = 10 hours

Question 10:Â Karan and Arjun run a 100-meter race, where Karan beats Arjun 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

a)Â Karan and Arjun reach the finishing line simultaneously.

b)Â Arjun beats Karan by 1 metre

c)Â Arjun beats Karan by 11 metres.

d)Â Karan beats Arjun by 1 metre.

Solution:

The speeds of Karan and Arjun are in the ratio 10:9. Let the speeds be 10s and 9s.
Time taken by Karan to cover 110 m = 110/10s = 11/s
Time taken by Arjun to cover 100 m = 100/9s = 11.11/s
Therefore, Karan reaches the finish line before Arjun. From the options, the only possible answer is d).

Question 11:Â Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

a)Â 3

b)Â 3.5

c)Â 4

d)Â 4.5

e)Â 5

Solution:

Let the distance be D.
Time taken by Arun = D/30
Time taken by Barun = D/40
Now, D/40 = D/30 – 2
=> 3D = 4D – 240
=> D = 240
Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr
Time Kiranmala takes to cover 240 km = 240/60 = 4 hr
So, Kiranmala has to start 4 hours after Arun.

Question 12:Â 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?

a)Â $\frac{7}{3}$

b)Â $\frac{4}{3}$

c)Â $\frac{5}{3}$

d)Â $\frac{8}{3}$

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 $24R + 12S – 24R + 12S = 4R^2 – S^2$
=> $24S = 6R^2 – 6S^2 and 24S = 4R^2 – S^2$
=> $6R^2 – 6S^2 = 4R^2 – S^2$
=> $2R^2 = 5S^2$
=> $24S = 10S^2 – S^2 = 9S^2$
=> $S = 24/9 = 8/3$

Question 13:Â Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyomâ€™s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?

a)Â 40

b)Â 50

c)Â 60

d)Â 80

Solution:

Let the number of steps on the escalator be x.

So, by the time Shyama covered 25 steps, the escalator moved ‘x-25’ steps.

Hence, the ratio of speeds of Shyama and escalator = 25:(x-25)

Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20)

But the ratio is 3:2

Ratio of speeds of Shyama and Vyom = 25(x-20)/20*(x-25) = 3/2

=> 10(x-20) = 12(x-25)

=> 2x = 100 => x = 50

Question 14:Â Thereâ€™s a lot of work in preparing a birthday dinner. Even after the turkey is in the oven, thereâ€™s still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table.
Three friends â€” Asit, Arnold and Afzal â€” work together to get all of these chores done. The time it takes them to do the work together is 6 hr less than Asit would have taken working alone, 1 hr less than Arnold would have taken alone, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?

a)Â 20 min

b)Â 30 min

c)Â 40 min

d)Â 50 min

Solution:

Let the time taken working together be t.
Time taken by Arnold = t+1
Time taken by Asit = t+6
Time taken by Afzal = 2t
Work done by each person in one day =Â $\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}$
Total portion of workdone in one day $=\frac{1}{t}$
$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$
$\frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{2-1}{2t}$
$2t+7=\frac{(t+1)\cdot(t+6)}{2t}$
$3t^2-7t+6=0Â \longrightarrow\ t=\frac{2}{3}$or $t=-3$
Therefore total time = $\frac{2}{3}$hours = 40mins

Alternatively,
$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$
From the options, if time $= 40$ min, that is, $t = \frac{2}{3}$
LHS =Â $\frac{3}{5} + \frac{3}{20} + \frac{3}{4} = \frac{(12+3+15)}{20} = \frac{30}{20} = \frac{3}{2}$
RHS = $\frac{1}{t}=\frac{3}{2}$
The equation is satisfied only in case of option C
Hence, C is correct

Question 15:Â On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend to the patient at the hospital.
Assume that a total ofÂ 1 min is elapsed for taking the patient into and out of the ambulance?

a)Â 4 min

b)Â 2.5 min

c)Â 1.5 min

d)Â The patient died before reaching the hospital

Solution:

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5

Hence, total time = 5 + 32.5 + 1 = 38.5 mins

So, the doctor would get 1.5 min to attend to the patient.

Question 16:Â Three small pumps and a large pump are filling a tank. Each of the three small pump works at 2/3 the rate of the large pump. If all four pumps work at the same time, they should fill the tank in what fraction of the time that it would have taken the large pump alone?
[CAT 2002]

a)Â 4/7

b)Â 1/3

c)Â 2/3

d)Â 3/4

Solution:

Let the work done by the big pump in one hour be 3 units.
Therefore, work done by each of the small pumps in one hour = 2 units.
Let the total work to be done in filling the tank be 9 units.
Therefore, time taken by the big pump if it operates alone = 9/3 = 3 hours.
If all the pumps operate together, the work done in one hour = 3 + 2*3 = 9 units.
Together, all of them can fill the tank in 1 hour.
Required ratio = 1/3

Question 17:Â In a 10 km race, A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. I f A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

Solution:

By the time A traveled 10 KM, B traveled 9 KM
Hence $Speed_A : Speed_B = 10:9$
Similarly $Speed_B : Speed_C = 10:9$
Hence $Speed_A : Speed_B : Speed_C = 100:90:81$
Hence by the time A traveled 10 KMs , C should have traveled 8.1 KMs
So A beat C by 1.9 KMs = 1900 Mts

Question 18:Â Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance traveled by him is

a)Â 55 miles

b)Â 60 miles

c)Â 65 miles

d)Â 70 miles

Solution:

Let the distance between the home and office be $2x$ miles
Time taken for going in the morning = $\frac{2x}{60}$ hrs
Time taken for going back in the evening = $\frac{x}{25} + \frac{x+5}{50}$. hrs
It is given that he took 30 minutes (0.5 hrs) more in the evening

Hence $\frac{2x}{60}$ hrs + 0.5 = $\frac{x}{25} + \frac{x+5}{50}$
Solving for x, we get x = 15 miles.
Total distance traveled = 2x + x + x + 5 = 4x + 5 = 65 Miles

Question 19:Â A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B7

a)Â 364

b)Â 378

c)Â 380

d)Â 388

Solution:

Let the distance traveled by the car be x KMs
Distance traveled by the bike = 168 KMs
Speed of car is double the speed of bike
=> $\frac{x}{3:40 – 2:00}$ = $2 \times \frac{168}{3:40 – 1:00}$
=> $\frac{x}{100}$ = $2 \times \frac{168}{160}$
=> x = 210
Hence the distance between A and B is x + 168 = 378 KMs

Question 20:Â Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Solution:

Let the distance between A and B be 4x.
Length of BP is thrice the length ofÂ AP.
=> AP = x and BP = 3x

Let the speed of car 1 be s and the speed of car 2 be 0.5s.
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.

=> x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60
x/s = 12

Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.

## CAT Time Speed And Distance Questions – FAQs

##### Is time speed and distance important for CAT?

Yes, Time Speed Distance is one of the important concepts from arithmetic in the Quantitative Aptitude section in CAT. A significant number of questions may appear in the CAT exam every year.

##### How do you solve time, speed and distance questions?

Basically, the questions from this topic are completely formula based. One must have to remember the basic formulas and can be able to understand the situation in the given problem. SomeÂ easy tips and tricks to solve the time speed and distance questions PDF. Download the PDF for free and be well-versed with the basics and the shortcuts to solving questions.