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# Time Speed and Distance Questions for CAT

Time Speed and Distance is a very important topic in the CAT Quant (Arithmetic) section. These questions are not very tough if you practice them well. Learn all the important Tricks and how to solve the questions on Time Speed and Distance. You can check out these CAT Time Speed and Distance questions from the In this post, we will look into some important Time Speed and Distance Questions for CAT Quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important CAT Time Speed and Distance Questions and answers PDF along with the detailed solutions below, which is completely Free.

Question 1: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 2: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 3: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 4: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 5: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 6: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 7:Â It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed?
[CAT 2002]

a)Â 6.40 pm

b)Â 7 pm

c)Â 7.20 pm

d)Â 8 pm

Solution:

Let the work done by each technician in one hour be 1 unit.
Therefore, total work to be done = 60 units.
From 11 AM to 5 PM, work done = 6*6 = 36 units.
Work remaining = 60 – 36 = 24 units.
Work done in the next 3 hours = 7 units + 8 units + 9 units = 24 units.
Therefore, the work gets done by 8 PM.

Question 8: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 9: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 10: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 11:Â In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.)

a)Â Akshay, $\frac{1}{12}$ mile

b)Â Chinmay, $\frac{1}{32}$ mile

c)Â Akshay, $\frac{1}{24}$ mile

d)Â Chinmay, $\frac{1}{16}$ mile

Solution:

Akshay can complete 1600 – 128 = 1472 m and Bhairav can completer 1600 m in the same time.

Bhairav can complete 100 m and Chinmay can complete 96 m in the same time.

=> Bhairav can complete 1600 m and Chinmay can complete 1536 m in the same time.

=> Akshay can complete 1472 m and Chinmay can complete 1536 m in the same time.

1.5 miles => 2400 m

Distance travelled by Akshay by the time Chinmay completes 1.5 miles = $\frac{1472}{1536}*2400$ = 2300 m

=> Akshay lost by 100 m, which is $\frac{1}{16}th$ of a mile.

Question 12: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 13: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 14: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 15: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.