CAT Time and Work Very Important Questions PDF [With Solutions]

Time and Work for CAT is an important topic in the CAT Quant section. Over the past few years, Time and Work has made a recurrent appearance in the Quant Section of the CAT. You can check out these Time and Work CAT Previous year questions. Practice a good amount of sums on CAT Time and Work questions. In this article, we will look into some important Time and Work Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT Time and Work Questions PDF below, which is completely Free.

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Question 1: 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) Karan and Arjun reach the finishing line simultaneously.

b) Arjun beats Karan by 1 metre

c) Arjun beats Karan by 11 metres.

d) Karan beats Arjun by 1 metre.

1) Answer (D)

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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: 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) 3

b) 3.5

c) 4

d) 4.5

e) 5

2) Answer (C)

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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: Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom’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) 40

b) 50

c) 60

d) 80

3) Answer (B)

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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 4: 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 of 1 min is elapsed for taking the patient into and out of the ambulance?

a) 4 min

b) 2.5 min

c) 1.5 min

d) The patient died before reaching the hospital

4) Answer (C)

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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

It takes 1 min to load and unload the patient.

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

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

Question 5: 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) 6.40 pm

b) 7 pm

c) 7.20 pm

d) 8 pm

5) Answer (D)

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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.

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Question 6: 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) 4/7

b) 1/3

c) 2/3

d) 3/4

6) Answer (B)

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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 7: 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) 25, 130

b) 30,150

c) 20, 90

d) None of these

7) Answer (D)

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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 8: 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) 12 hours

b) 10 hours

c) 8 hours

d) 6 hour

8) Answer (B)

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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 9: 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) 1 hour

b) 50 minutes

c) 1/2 hour

d) 55 minutes

9) Answer (D)

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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 10: 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) Akshay, $\frac{1}{12}$ mile

b) Chinmay, $\frac{1}{32}$ mile

c) Akshay, $\frac{1}{24}$ mile

d) Chinmay, $\frac{1}{16}$ mile

10) Answer (D)

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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.

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