# Arithmetic - Time & Work - Previous Year CAT/MBA Questions

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**CAT 2023 QA Slot 1 | Arithmetic - Time & Work CAT Question**

The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is

Answer: 27

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**Text Explanation** :

Let the time taken by A and K to complete a work is x and 2x days respectively.

Work done in a day is the efficiency of a person.

Hence, if efficiencies of A, S and K are in Harmonic Progression, time taken by them to finish a work will be in Arithmetic Progression.

∴ Time taken by S alone is arithmetic mean of time taken by A and K alone.

⇒ Time taken by S alone = (x + 2x)/2 = 1.5x

Using unitary method

⇒ 1 = $\frac{1}{\mathrm{x}}\times 4$ + $\frac{1}{1.5\mathrm{x}}\times 9$ + $\frac{1}{2\mathrm{x}}\times 16$

⇒ x = 4 + 6 + 8

⇒ x = 18

∴ S will take 1.5 × 18 = 27 days to finish the job alone.

Hence, 27.

**Concept**:

Workspace:

**CAT 2023 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

- (a)
60

- (b)
90

- (c)
75

- (d)
120

Answer: Option B

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**Text Explanation** :

Let time taken by A and C alone to fill the tank is A and C hours respectively.

∴ Time taken B alone = (A - 1) hours.

**Case 1**: All three can fill the tank in 2 hours.

⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$ ...(1)

**Case 2**: B works for 1 hour and C works for 5/4 hours.

⇒ - $\frac{1}{\mathrm{A}-1}\times 1$ + $\frac{1}{\mathrm{C}}\times \frac{9}{4}$ = 1 ...(2)

(1) × 9/4 - (2)

⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{9}{4(\mathrm{A}-1)}$ + $\frac{1}{\mathrm{A}-1}\times 1$ = $\frac{9}{8}$ - 1

⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{5}{4(\mathrm{A}-1)}$ = $\frac{1}{8}$

⇒ $\frac{18}{\mathrm{A}}$ - $\frac{10}{(\mathrm{A}-1)}$ = 1

⇒ A(A - 1) = 18(A - 1) - 10A

⇒ A^{2} - A = 8A - 18

⇒ A^{2} - 9A + 18 = 0

⇒ A = 3 or 6 hours.

Since A should be less than 5, hence we accept only A = 3 hours.

Now, from (1), we have

⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ $\frac{1}{3}$ - $\frac{1}{2}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ C = 3/2 hours = 90 minutes.

Hence, option (b).

**Concept**:

Workspace:

**CAT 2023 QA Slot 3 | Arithmetic - Time & Work CAT Question**

Rahul, Rakshita and Gurmeet, working together, would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be

- (a)
21

- (b)
17

- (c)
16

- (d)
20

Answer: Option A

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**Text Explanation** :

Let the total work to be done = LCM(7, 15, 6) = 630 units.

⇒ Combined efficiency of Rahul, Rakshita and Gurmeet < 630/7 = 90 units/day

All 3 worked for 6 days and then Rakshita worked for 3 days. Calculating total work done

∴ 630 = (combined efficiency of all 3) × 6 + (efficiency of Rakshita) × 3

⇒ (efficiency of Rakshita) × 3 = 630 - (combined efficiency of all 3) × 6

⇒ (efficiency of Rakshita) = 210 - (combined efficiency of all 3) × 2

⇒ (efficiency of Rakshita) > 210 - 90 × 2

⇒ (efficiency of Rakshita) > 30

∴ Rakshita's efficiency is greater than 30 units/day.

⇒ Time taken by Rakshita < 630/30 = 21 days.

∴ Rakshita will take less than 21 days to finish the job. Hence, she cannot take 21 days to finish the job.

Hence, option (a).

Workspace:

**CAT 2023 QA Slot 3 | Arithmetic - Time & Work CAT Question**

Gautam and Suhani, working together, can finish a job in 20 days. If Gautam does only 60% of his usual work on a day, Suhani must do 150% of her usual work on that day to exactly make up for it. Then, the number of days required by the faster worker to complete the job working alone is

Answer: 36

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**Text Explanation** :

Let work done per day (efficiency) of Gautam and Suhani are 'g' and 's' units.

A shortfall of 40% for Gautam is compensated by 50% extra work done by Suhani.

⇒ 0.4 × g = 0.5 × s

⇒ g/s = 5/4

⇒ Let g = 5x and s = 4x

Together they can complete the work in 20 days.

⇒ Total work to be done = 20 × (5x + 4x) = 180x units

The faster among the two is Gautam whose efficiency is 5x.

∴ Time required by Gautam alone = 180x / 5x = 36 days.

Hence, 36.

Workspace:

**CAT 2022 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is:

Answer: 6

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**Text Explanation** :

Ratio of time taken by Anu, Tanu and Manu is 5 : 8 : 10.

⇒ Ratio of efficiencies of Anu, Tanu and Manu = $\frac{1}{5}$ : $\frac{1}{8}$ : $\frac{1}{10}$ = 8 : 5 : 4

Let their efficiencies be 8x, 5x and 4x respectively per hour.

Total work done by them in 4 days = (8x + 5x + 4x) × 4 × 8 = 17x × 32 = 544x

Now, Anu and Tanu worked by 6 days working 6 hours 40 minute i.e., 6$\frac{2}{3}$ hours daily.

∴ Worked completed by Anu and Tanu = 13x × 6 × $\frac{20}{3}$ = 520x

⇒ Time taken by Manu to complete the remaining work = $\frac{544x-520x}{4x}$ = 6 hours.

Hence, 6.

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Time & Work CAT Question**

A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of N is

- (a)
23

- (b)
140

- (c)
36

- (d)
150

Answer: Option B

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**Text Explanation** :

Let the efficiency of each person be ‘e’ units/hour.

Given,

35% work is done by N men in 10 days working for 7 hours/day, while

65% work is done by N-10 men in 14 days working for 10 hours/day

⇒ $\frac{\mathrm{N}\times 10\times 7}{0.35}$= $\frac{(\mathrm{N}-10)\times 14\times 10}{0.65}$

⇒ $\frac{\mathrm{N}\times 7}{7}$= $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ N = $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ 13N = 14N – 140

⇒ N = 140

Hence, option (b).

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Time & Work CAT Question**

Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is

Answer: 11

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**Text Explanation** :

Bob takes 40 days to finish the work. Alex takes is twice as fast as Bob and hence will take half the time taken by Bob i.e., 20 days. Alex is also thrice as fast as Cole, hence Cole will take thrice the time taken by Alex, i.e., 60 days.

Time take by

Alex = 20 days

Bob = 40 days

Cole = 60 days

Let the total work to be done = 120 units.

∴ Efficiency of

Alex = 6 units/day

Bob = 3 units/day

Cole = 2 units/day

Work done/cycle = 22 units.

∴ Work done in 5 cycles = 5 × 22 = 110 units

Work left after 5 cycles (15 days) = 120 – 110 = 10 units.

On 16th day Alex and Bob will together complete 9 units of work while remaining 1 unit of work will be completed by Bob and Cole on 17th day.

∴ Alex worked for 10 days in 5 complete cycles + on 16th day i.e., total 11 days.

Hence, 11.

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Time & Work CAT Question**

Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is

- (a)
8

- (b)
6

- (c)
5

- (d)
7

Answer: Option D

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**Text Explanation** :

Let the total work to be done = LCM (15, 12, 20) = 60 units.

Efficiency of Anu = 60/15 = 4 units/day

Efficiency of Vinu = 60/12 = 5 units/day

Efficiency of Manu = 60/20 = 3 units/day

Let us calculate the work done every 2 days.

Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units

Anu works on alternate days, hence work done by Anu = 4 units

Manu works on alternate days, hence work done by Manu = 3 units

∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.

Hence total work done in 6 days = 3 × 17 = 51 units.

On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.

∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]

∴ The number of days needed to complete the work is 7 days.

Hence, option (d).

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Time & Work CAT Question**

Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

Answer: 32

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**Text Explanation** :

Let the total work to be done = LCM (12, 16, 24) = 48 units.

Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month

⇒ x + y = 48/12 = 4 …(2)

⇒ y + z = 48/16 = 3 …(1)

⇒ z + x = 48/24 = 2 …(3)

Adding all these equations

⇒ x + y + z = 9/2 = 4.5 …(4)

Solving these four equations we get,

x = 1.5, y = 2.5 and z = 0.5

∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.

∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.

Hence, 32.

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is

- (a)
9100

- (b)
9000

- (c)
9150

- (d)
9200

Answer: Option A

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**Text Explanation** :

Let the total work to be done = LCM (60, 84) = 420 units.

Efficiency of Anil = 420/60 = 7 units/day

Efficiency of Bimal = 420/84 = 5 units/day

Work done by Anil in 10 + 14 days = 24 × 7 = 168 units

Work done by Bimal in 14 days = 14 × 5 = 70 units

∴ Work done by Charu = 420 – 70 - 168 = 182 units

Fraction of work done by Charu = 182/420 = 91/210

∴ Payment received by Charu = 91/210 × 21,000 = Rs. 9,100

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is:

- (a)
140

- (b)
120

- (c)
144

- (d)
264

Answer: Option C

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**Text Explanation** :

Let the filling and emptying capacity of A and B be ‘a’ and ‘b’ units/hour respectively.

**Case 1**: A is opened at 2 pm and B at 3 pm

Total work done till 10 pm = 8a - 7b

**Case 2**: A is opened at 2 pm and B at 4 pm

Total work done till 6 pm = 4a - 2b

Since work done is same in both cases, we have

8a – 7b = 4a – 2b

⇒ 4a = 5b

Now time taken by A alone to fill the tank = (Total work)/a = (4a – 2b)/a = (5b – 2b)/(5b/4) = 12/5 hours = 144 minutes.

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 3 | Arithmetic - Time & Work CAT Question**

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is:

- (a)
10

- (b)
12

- (c)
12.5

- (d)
11.5

Answer: Option A

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**Text Explanation** :

Let the efficiency of Rahul be ‘r’ units per hour and for Gautam be ‘g’ units per day.

Initially, Rahul works for 8 hours and Gautam works for 6 hours.

∴ Work done by them = 8r + 6g …(1)

If both of them worked starting from 9 AM, they would have completed the work in 7.5 hours.

∴ Work done by them = 7.5r + 7.5g …(2)

⇒ 8r + 6g = 7.5r + 7.5g

⇒ 0.5r = 1.5g

⇒ r = 3g

∴ Work to be done = 8r + 6g = 8r + 2r = 10r

∴ Time taken by Rahul to complete the work alone = 10r/r = 10 hours.

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 3 | Arithmetic - Time & Work CAT Question**

Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

Answer: 3000

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**Text Explanation** :

Let the area to be painted = LCM(12, 16) = 48 units.

⇒ Efficiency of Anil = 48/12 = 4 units/day

and Efficiency of Barun = 48/16 = 3 units/day

Work done by Anil in 6 days = 6 × 4 = 24 units

Work done by Anil in 6 days = 6 × 3 = 18 units

∴ Remaining work done by Chandu = 48 – 24 – 18 = 6 units.

⇒ Payment received by Chandu = 24000/48 × 6 = Rs. 3,000

**Alternately,**

Fraction of work done by Anil in 6 days = 6/12 = ½

Fraction of work done by Barun in 6 days = 6/16 = 3/8

⇒ Fraction of work done by Chandu in 6 days = 1 - ½ - 3/8 = 1/8

Since, Chandu completes 1/8^{th} of the work, he will receive 1/8^{th} of the payment.

∴ Chandu’s payment = 1/8 × 24,000 = Rs. 3,000

Hence, 3000.

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time & Work CAT Question**

John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

Answer: 4

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**Text Explanation** :

Let John’s efficiency be 1 unit/day.

∴ Jack’s efficiency is 2 units/day.

Jack and Jim’s efficiency is thrice John’s efficienc.

∴ 2 + e_{Jim} = 3 × 1

⇒ e_{Jim} = 1

John takes three days more than that taken by three of them working together

Let the time taken by all three together is d.

∴ Total work to be done = 4 × d = 1 × (d + 3)

⇒ d = 1

∴ John finished the work in d + 3 = 4 days.

Since, John and Jim have same efficiency, Jim will also finish the work in 4 days.

Hence, 4.

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Time & Work CAT Question**

A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?

Answer: 40

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**Text Explanation** :

In 60 days 1.5 kms out of 6 kms is built.

∴ In 60 days 1.5/6 = 1/4^{th} work is done.

To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.

Days remaining now is 200 – 60 = 140 days.

∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.

The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.

∴ 180 × 140 = 140 × (140 + x)

⇒ x = 40

∴ Contractor hires 40 more people.

Hence, 40.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time & Work CAT Question**

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

- (a)
36

- (b)
24

- (c)
12

- (d)
18

Answer: Option D

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**Text Explanation** :

Let total work be the LCM of 12 and 9 = 36 units.

Let the efficiency of A and B be a and b respectively.

Work done per day when A and B are working together = 36/12 = 3 units.

∴ a + b = 3 ...(1)

Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.

∴ (a/2) + 3b = 4 ...(2)

Solving (1) and (2), we get; a = 2.

Time taken by A alone to complete the work = 36/2 = 18 days.

Hence, option (d).

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time & Work CAT Question**

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

Answer: 13

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**Text Explanation** :

Let the work done by one man and one machine per day be x and y respectively.

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.

Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.

∴ (3x + 8y) = 2(8x + 3y)

∴ 13x = 2y.

So, work done by 13 men in a day = work done by 2 machines in a day.

∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.

Hence, 13.

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

- (a)
13

- (b)
12

- (c)
15

- (d)
14

Answer: Option A

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**Text Explanation** :

Anil in one day can do 1/20^{th} of the work

Sunil in one day can do 1/40^{th} of the work

Anil starts the job and Sunil joins him after three days.

So, Anil would have done 3/20^{th} of the work by the time Sunil joins

After Sunil joins, they both would be doing 3/40^{th} of work everyday

Now, it is known that Bimal joins them after some days and finishes 10% of the work (i.e. 1/10^{th }of the work).

Now, Anil alone had done 3/20^{th} of the work in first 3 days and Bimal completes 1/10^{th} of the work

So, in total they would have done 3/20 + 1/10 = 1/4^{th} of the work.

Remaining work = 3/4^{th}, which would be done by Anil and Sunil together.

Anil and Sunil together comple 1/20^{th} + 1/40^{th} = 3/40^{th }work in 1 day.

Hence, time taken by them to comple 3/4^{th} of the work = 10 days.

Combining everything,

Total number of days = 3 + 10 = 13 days.

Hence, option (a).

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time & Work CAT Question**

John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

Answer: 12

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**Text Explanation** :

If John works the same number of regular and over-time hours say 'p'

The income would be 57p and 114p

Let's say that he works 'x' hours regular and 'y' hours overtime...

So, the income would be 57x and 114y

we are told that 114y is 15% of 57x

114y = 0.15 × 57x

y = 0.075x

we also know that x + y = 172

therefore, x + 0.075x = 1.075x = 172

x = 160

y = 172 - 160 = 12

Therefore, the number of hours he worked overtime is 12 hours.

Hence, 12.

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work CAT Question**

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Answer: 10

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**Text Explanation** :

Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.

Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b)

Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)

6(6a – 5b) = 60(5a – 6b)

∴ 6a – 5b = 50a – 60b

∴ 44a = 55b

∴ a = 5b/4

Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.

Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)

∴ m(2a – b) = 6(6a – 5b)

∴ $\mathrm{m}\left(2\times \frac{5}{4}\mathrm{b}-\mathrm{b}\right)$ = $6\left(6\times \frac{5}{4}\mathrm{b}-5\mathrm{b}\right)$

∴ $\mathrm{m}\left(\frac{6\mathrm{b}}{4}\right)$ = $6\left(\frac{10\mathrm{b}}{4}\right)$

Solving this, we get m = 10

Hence, 10.

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work CAT Question**

Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

- (a)
36

- (b)
32

- (c)
45

- (d)
40

Answer: Option B

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**Text Explanation** :

Let work-done by a human and a robot in a day are ‘h’ and ‘r’ respectively.

Work done by 15 human and 5 robot in 30 days = 30(15h + 5r)

Work done by 5 human and 15 robot in 60 days = 60(5h + 15r)

∴ Total work = 30(15h + 5r) = 60(5h + 15r)

⇒ 15h + 5r = 10h + 30r

⇒ h = 5r

Let fifteen humans take ‘y’ days to finish the job.

Work done by 15 humans in y days = y × 15

∴ 30(15h + 5r) = y × 15h

⇒ 30(15h + h) = y × 15h

⇒ y = 32

Hence, option (b).

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work CAT Question**

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

- (a)
18

- (b)
22

- (c)
16

- (d)
20

Answer: Option D

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**Text Explanation** :

Ratio of time taken by A and B to complete a certain task = 4 : 5.

∴ Ratio of their efficiencies = 5 : 4.

Now, A completes 50% of the work alone and B completes 5% of the work alone. That means, A and B together complete 45% of the work together in 4 days.

Out of this 45%, work done by A and B individually will be in the ratio of their efficiencies.

∴ Work done by B in 4 days = 4/9^{th} of 45% = 20%. [Since the ratio of their efficiencies is 5 : 4]

⇒ B completes 20% (i.e., 1/5^{th}) of the work in 4 days.

Thus, B alone can finish the entire job in 20 days.

Hence, option (d).

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work CAT Question**

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Answer: 48

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**Text Explanation** :

Suppose the inlet pipes of type A fill in water at the rate ‘a’ units per minute and the inlet pipes of type B fill in water at the rate ‘b’ units per minute.

Therefore we have the following

30(10a + 45b) = 60(8a + 18b)

∴ 300a + 1350b = 480a + 1080b

∴ 180a = 270b

∴ a = 1.5b

Total capacity of the tank

= 300a + 1350b = 300(1.5b) + 1350b

= 1800b

If 7 inlet pipes of type A and 27 inlet pipes of type B are opened, the volume of water filled in every minute

= 7a + 27b = 7(1.5b) + 27b = 37.5b

Therefore the number of minutes taken to fill the tank

$=\frac{\mathrm{1800b}}{37.\mathrm{5b}}=\mathrm{48\; minutes}$

Hence, 48.

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work CAT Question**

Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his efficiency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?

- (a)
12

- (b)
11

- (c)
13.5

- (d)
14.5

Answer: Option C

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**Text Explanation** :

Suppose ‘R’ and ‘G’ are the number of units of work completed by Ramesh and Ganesh everyday. Therefore the total work to be completed = 16(R + G) = 16R + 16G

For the first 7 days, both work at 100% efficiency and for the remaining 10 days, Ramesh works at 70% efficiency. Therefore the total work to be completed = 7(R + G) + 10(0.7R + G) = 14R + 17G

∴ 16R + 16G = 14R + 17G or G = 2R

Now if only Ganesh had worked after Ramesh fell sick:

Work to be done = 9(G + R) = 9(2R + R) = 27R

Efficiency of Ganesh = G = 2R

∴ Time taken to finish the remaining work = 27R/2R = 13.5 days.

Hence, option (c).

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work CAT Question**

A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

- (a)
4:36 pm

- (b)
4:12 pm

- (c)
4:48 pm

- (d)
4:24 pm

Answer: Option D

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**Text Explanation** :

The time taken by pump A alone to fill the tank completely = t hours.

The time taken by pump B alone to fill the tank completely = ‘t - 2’ hours.

On Wednesday, pump A was used for 3 hour less and pump B was used for 2 hours to complete the remaining work.

⇒ Work done by A in 3 hours = Work done by B in 2 hours.

⇒ efficiency of A : efficiency of B = 2 : 3

Let efficiency of A be '2x' and that of B be '3x'.

Now, total work is done by B in (t - 2) hours and by A in t hours.

⇒ 3x × (t - 2) = 2x × t

⇒ 3(t - 2) = 2t

⇒ t = 6 hours

∴ The tank starts filling at 2 pm.

If both pumps start filling, their combined efficiency = 5x

Total work to be done = 3x(t - 2) = 12x

⇒ Time taken to fill the tank = 12x/5x = 2.4 hours = 2 hours 24 minutes.

Therefore on Thursday, the tank will be completely full at 4.24 pm.

Hence, option (d).

Workspace:

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