# CAT Time and Work Questions PDF [Important]

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: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

**1)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 2: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

**2)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 3: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

**3)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.

**Question 4: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

**4)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 5: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

**5)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.

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**Question 6: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

**6)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 7: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

**7)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)

**Question 8: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

**8)Â AnswerÂ (B)**

**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 9: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

**9)Â AnswerÂ (B)**

**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 10:Â **Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal’s share, in rupees, is

a)Â 100

b)Â 200

c)Â 300

d)Â 400

**10)Â AnswerÂ (A)**

**Solution:**

Let the time take by kamal to complete the task be x days.

Hence we have $\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$

=> x = 40 days.

Ratio of the work done by them = $\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$ = 4 : 5 : 1

Hence the wage earned by Kamal = 1/10 * 1000 = 100