# CAT Time and Work Questions PDF [Most Important]

Time and Work is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Time and Work questions in the **CAT Previous year’s papers. **If you want to learn the basics, you can watch these videos on **Time and Work basics**. This article will look into some important Time and Work Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Time and Work Most Important Questions PDF below, which is completely Free.

Download Time and Work Questions for CAT

**Question 1: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 $\frac{8}{3}$

**1)Â AnswerÂ (D)**

**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 2: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

**2)Â AnswerÂ (B)**

**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 3: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

**3)Â AnswerÂ (C)**

**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 4: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

**4)Â AnswerÂ (C)**

**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: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

**5)Â AnswerÂ (D)**

**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.

Checkout: CAT Free Practice Questions and Videos

**Question 6: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

**6)Â AnswerÂ (B)**

**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: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

**7)Â AnswerÂ (B)**

**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 8: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

**8)Â AnswerÂ (D)**

**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 9: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

**9)Â AnswerÂ (A)**

**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 10: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

**10)Â AnswerÂ (C)**

**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)