◆ QA · Arithmetic

Time & Work · Pipes & Cisterns, formulas + CAT PYQs

The highest-yield arithmetic family in CAT. Master the LCM-of-work method once and you can crack work-rate, efficiency, wage-share, man-days and pipes-and-cisterns problems with simple arithmetic, no algebra-heavy setups needed.

16formulas

35CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Formula & Concept Sheet

A-to-Z. Everything you need for Time & Work and Pipes & Cisterns, distilled.

1Work = Rate × Time

The master identity. Work done equals how fast you go times how long you go.
Rate (efficiency) and time are inversely proportional for a fixed amount of work.

Work = Rate × Time

2Per-day work (the 1/x idea)

If A finishes a job in x days, then in 1 day A does 1/x of the job.
Conversely, if A does 1/x of a job in a day, A needs x days for the whole job.

1 day's work = 1/x

3Combined rate (working together)

Rates simply add when people/pipes work together.
Two workers (x & y days alone) finish together in xy/(x+y) days.

1/T = 1/a + 1/b + 1/c …

4LCM-of-work method (the big shortcut)

Set total work = LCM of the given times. This kills all fractions.
Each worker's efficiency = total work ÷ their solo time, in clean integer units/day.
A in x days, B in y days ⇒ take work = LCM(x, y); efficiency ratio A : B = y : x.

Work = LCM(x, y); Eff = Work/time

5Efficiency comparison

"A is twice as fast as B" ⇒ A's efficiency = 2 × B's; A takes half the time.
Efficiency is the inverse of time: faster worker ⇒ fewer days.

Eff_A / Eff_B = Time_B / Time_A

6Wages / payment share

Money is split in the ratio of work actually done (= efficiency × days worked).
If all work equal days, share simplifies to the ratio of efficiencies.

Share ∝ (efficiency × days worked)

7Man-days / chain rule

With M men, D days, H hours/day producing work W, the quantity below is constant.
Use it to relate two scenarios with different men, hours, days and amount of work.

(M × D × H) / W = constant M₁D₁H₁/W₁ = M₂D₂H₂/W₂

8A in x, B in y → together

Direct formula for two workers finishing a single job together.
Extends to subtraction when one is undoing work (a drain pipe).

T = xy / (x + y)

9Alternate-day / roster working

Find work done in one full cycle (e.g. 2-day or 3-day pattern), then count whole cycles.
Handle the leftover work day-by-day at the end, the job often finishes mid-cycle.

Days ≈ (Total work ÷ cycle work) × cycle length

10Pipes & Cisterns = Time & Work

Exactly the same analogy: a pipe "does work" by filling/emptying a tank.
Inlet = a pipe that fills (positive rate). Outlet/drain = empties (negative rate).

Inlet fills in x hrs ⇒ 1 hr fills 1/x

11Inlet + outlet together (leak)

Inlet fills in x hrs, outlet empties in y hrs ⇒ net fill in 1 hr is the difference.
If y > x the tank still fills; if y < x it can never fill.

Net/hr = 1/x − 1/y; Time = xy/(y − x)

12Two inlets together

Two filling pipes (x and y hrs) opened together fill faster, rates add.

Net/hr = 1/x + 1/y; Time = xy/(x + y)

13Two inlets + one drain

Pipes fill in x, y hrs; drain empties in z hrs; all open ⇒ add inlets, subtract drain.

Net/hr = 1/x + 1/y − 1/z Time = 1 / (1/x + 1/y − 1/z)

14Tank-capacity (LCM) trick

Assume tank capacity = a convenient number (often LCM of the times, e.g. 40 L).
Compute each pipe's flow in litres/hr, turns fractions into clean arithmetic.

Capacity = LCM(times); Flow = Cap/time

15Mixed pipe groups (solve as equations)

When several pipes of types A and B run, set total work = (net efficiency) × time.
Equate two scenarios to find the ratio of efficiencies, then plug into a third.

Cap = (n_A·a + n_B·b) × time

16Clocks (bonus, same chapter topic)

Minute and hour hands coincide every 720/11 ≈ 65.45 min.
The hands meet 22 times in a 24-hour day; a fast clock meets more often.

Hands meet every 720/11 min

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

CAT Previous-Year Questions

Real CAT questions with worked solutions from the book. Difficulty: Easy Moderate Hard. Click any to reveal the solution.

CAT 1997

Moderate CAT 1997

Direction (Q. 1 and 2): Answer the questions based on the following information.

The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.

1. Raja starts working on February 25, 1996, and finishes the job on March 2, 1996. How much time would T and J take to finish the same job if both start on the same day as Raja?

(1) 4 days
(2) 5 days
(3) Either (1) or (2)
(4) Cannot be determined

Show solution

(1) 4 days. No day of the week begins with "R", so Raja never gets a holiday: from Feb 25 to Mar 2, 1996 he works 7 days, so total work = 7 units at 1 unit/day. T is off on Tuesday and Thursday; no day begins with "J", so J works every day. Feb 25, 1996 is a Sunday, so the combined T + J work runs Sun = 1, Mon = 2, Tue = 1, Wed = 2 … The 7 units are completed in 4 days.

Hard CAT 1997

2. Starting on February 25, 1996, if Raja had finished his job on April 2, 1996, when would T and S together have completed the job, had they started on the same day as Raja?

(1) March 15, 1996
(2) March 14, 1996
(3) March 22, 1996
(4) Data insufficient

Show solution

(3) March 22, 1996. Raja works Feb 25 → Apr 2 = 38 days, so total work = 38 units at 1 unit/day. T is off on Tue & Thu; S is off on Sat & Sun. Their combined pattern gives 10 units per week, so 30 units in three weeks. The remaining 8 units, starting from Sunday Feb 25, are completed by March 22, 1996.

CAT 2001

Hard CAT 2001

A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair took two-third the time needed by the second pair to complete the work. Which is the first pair?

(1) A, B
(2) A, C
(3) B, C
(4) A, D

Show solution

(4) A, D. Take work = 4 units. Efficiencies: A = 1, B = ½, C = ¼, D = ⅛ unit/day. The faster pair has 3/2 the efficiency of the slower pair (since it takes ⅔ the time). (A + D) = 1 + ⅛ = 9/8; (B + C) = ½ + ¼ = 6/8. And 9/8 = (3/2) × 6/8. So the first (faster) pair is A, D.

Moderate CAT 2001

There's a lot of work in preparing a birthday dinner. Even after the turkey is in oven, there are still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. Three friends, Asit, Arnold, and Afzal, work together to get all of these chores done. The time it takes them to do the work together is six hours less than Asit would have taken working alone, one hour less than Arnold would have taken, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?

(1) 20 minutes
(2) 30 minutes
(3) 40 minutes
(4) 50 minutes

Show solution

(3) 40 minutes. Let t = time together. Then 1/(t+6) + 1/(t+1) + 1/(2t) = 1/t. Solving gives 3t² + 7t − 6 = 0 ⇒ t = ⅔ hr = 40 minutes.

CAT 2002

Moderate CAT 2002

It takes 6 technicians a total of 10 hours 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 a.m. and one technician per hour is added beginning at 5 p.m., at what time will the server be completed?

(1) 6:40 p.m.
(2) 7:00 p.m.
(3) 7:20 p.m.
(4) 8:00 p.m.

Show solution

(4) 8:00 p.m. Each technician = 1 unit/hr, so total work = 6 × 10 = 60 units. From 11 a.m. to 5 p.m. (6 hrs) 6 workers do 36 units; 24 units remain. 5-6 p.m.: 7 units; 6-7 p.m.: 8 units; 7-8 p.m.: 9 units. 7 + 8 + 9 = 24, completed at 8 p.m.

CAT 2004

Hard CAT 2004

In Nuts and Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts. Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?

(1) 130 minutes
(2) 135 minutes
(3) 170 minutes
(4) 180 minutes

Show solution

(3) 170 minutes. Nuts: 1000 nuts take 10 min + 5 min clean = 15 min/batch; 9000 nuts = 9 batches = 135 min, but the final cleaning is unneeded ⇒ 130 min. Bolts: 1500 bolts take 20 min + 10 min clean = 30 min/batch; 9000 bolts = 6 batches = 180 − 10 = 170 min. The slower line governs ⇒ 170 minutes.

CAT 2017

Moderate CAT 2017 TITA

A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

Show solution

15 days. Each person = 1 unit/day, total work = 120. Work done on days 1, 2, 3 … = 1, 2, 3 … So cumulative = n(n+1)/2 = 120 ⇒ n = 15. The job finishes in 15 days.

Easy CAT 2017

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 ₹1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal's share, in rupees, is

(1) 100
(2) 200
(3) 300
(4) 400

Show solution

(1) ₹100. Take work = 40 units. Amal = 4, Bimal = 5, all-three = 40/4 = 10 units/day ⇒ Kamal = 10 − 4 − 5 = 1 unit/day. Shares are in ratio 4 : 5 : 1, so Kamal gets 1/10 × 1000 = ₹100.

CAT 2018

Moderate CAT 2018

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?

(1) 45
(2) 36
(3) 32
(4) 40

Show solution

(3) 32 days. (15h + 5r)·30 = (5h + 15r)·60 ⇒ 15h + 5r = 10h + 30r ⇒ 5h = 25r ⇒ h = 5r. Total work = (15h + 5r)·30 = (15h + h)·30 = 480h. Time for 15 humans = 480h/15h = 32 days.

Moderate CAT 2018

When they work alone, B needs 25% more time to finish a job than A does. They 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?

(1) 20
(2) 22
(3) 16
(4) 18

Show solution

(1) 20 days. Let A need 16a days, so B needs 20a days. A does half the job in 8a days; B does the final 5% in 20a × 0.05 = a days. So 8a + 4 + a = 13 ⇒ a = 1. Hence B alone needs 20a = 20 days.

Hard CAT 2018

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?

(1) 14.5
(2) 11
(3) 13.5
(4) 12

Show solution

(3) 13.5 days. Total = 16(R + G). The work normally done in 9 days got done in 10 days at reduced efficiency: 9(R + G) = 10(0.7R + G) ⇒ G = 2R. Remaining work when Ramesh fell sick = 9(R + G) = 9(0.5G + G) = 13.5G. Ganesh alone: 13.5G / G = 13.5 days.

Hard CAT 2018 TITA

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?

Show solution

13 men. Let a man = a, machine = b per day. 3a + 8b = 2(3b + 8a) ⇒ 13a = 2b. So the daily work of 13 men = the daily work of 2 machines. Since 2 machines finish in 13 days, 13 men also finish in 13 days ⇒ 13 men.

Moderate CAT 2018 TITA

John gets ₹57 per hour of regular work and ₹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?

Show solution

12 hours. Let regular hours = a, overtime = 172 − a. Overtime income = 15% of regular: (172 − a)·114 = 0.15·a·57 ⇒ a = 160. Overtime hours = 172 − 160 = 12 hours.

Moderate CAT 2018

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?

(1) 12
(2) 13
(3) 15
(4) 14

Show solution

(2) 13 days. Take work = 40 units; Anil = 2/day, Sunil = 1/day. Anil's first 3 days = 6 units. Bimal does 10% = 4 units. Remaining 40 − 6 − 4 = 30 units done by Anil + Sunil (3 units/day) in 30/3 = 10 days. Total = 3 + 10 = 13 days.

CAT 2020

Hard CAT 2020 TITA

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?

Show solution

4 days. Efficiencies: John : Jack = 1 : 2; Jack + Jim = 3 × John = 3, so Jim = 3 − 2 = 1. All three together = 1 + 2 + 1 = 4. Time ratio John : all-three = 4 : 1, and John = three-days-more: 4x − x = 3 ⇒ x = 1, so John = 4 days. Jim's efficiency equals John's, so Jim alone = 4 days.

CAT 2021

Hard CAT 2021 · Slot 1 TITA

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

Show solution

32 months. Let solo times be x, y, z months. 1/x + 1/y = 1/12; 1/y + 1/z = 1/16; 1/x + 1/z = 1/24. From these, 1/y = 9/96 − 1/24 = 5/96 (Akbar, fastest) and 1/z = 1/96 (Anthony, slowest). So Amar's rate = 1/x = 1/12 − 5/96 = 3/96, giving x = 96/3 = 32 months. Amar is neither fastest nor slowest ⇒ 32 months.

Hard CAT 2021 · Slot 2

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 ₹, proportionate to the work done by him, is

(1) 9100
(2) 9150
(3) 9000
(4) 9200

Show solution

(1) ₹9100. Anil's first 10 days = 10/60 = 1/6. Remaining 5/6 done by all three in 14 days: 14(1/60 + 1/84 + 1/x) = 5/6 ⇒ Charu alone = 420/13 days. Charu's work in 14 days = 14 ÷ (420/13) = 13/30. Share = (13/30)·21000 = ₹9100.

CAT 2022

Hard CAT 2022 · Slot 2 TITA

Working alone, the time 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:

Show solution

6 hours. Solo times 5x : 8x : 10x ⇒ hourly efficiencies in ratio 8 : 5 : 4. All three working 4 days × 8 hrs = 32 hrs finish the job, so total work = (8 + 5 + 4) × 32 = 544 units. Anu + Tanu work 6 days × 6 hrs 40 min = 40 hrs ⇒ (8 + 5) × 40 = 520 units. Remaining = 544 − 520 = 24 units, done by Manu at 4 units/hr ⇒ 24 ÷ 4 = 6 hours.

Hard CAT 2022 · Slot 3 TITA

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, 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:

Show solution

11 days. Rates: Bob = 3, Alex = 6, Cole = 2 units/day. Total = 40 × 3 = 120 units. Cycle: day1 = 9, day2 = 5, day3 = 8 ⇒ 22 units/3-day cycle. After 5 cycles (15 days) = 110 units. Day 16 (Alex + Bob) adds 9 ⇒ 119; finishes on day 17. Alex works on 2 of every 3 days = 2 × 5 + 1 (day 16) = 11 days.

Moderate CAT 2022 · Slot 3

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:

(1) 150
(2) 36
(3) 140
(4) 23

Show solution

(3) 140. Work done = N × 7 × 10 = 70N man-hours for 35%. Total work = 70N / 0.35 = 200N; remaining = 130N. Then 130N = (N − 10) × 14 × 10 ⇒ N = 140.

Moderate CAT 2022 · Slot 3 TITA

Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

Show solution

60 seconds. Let his speed = x steps/s, escalator = y steps/s. Total steps: 30(x + y) = 20(2x + y) ⇒ x = y. Total = 30(2y) = 60y. Standing still ⇒ time = 60y / y = 60 s.

CAT 2023

Hard CAT 2023 · Slot 1 TITA

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

Show solution

27 days. Work amounts in HP ⇒ times in AP. Kamal = 2 × Amal's time, so times ratio = 2 : 3 : 4 ⇒ work ratio = 6 : 4 : 3. Total work = 4·6 + 9·4 + 16·3 = 24 + 36 + 48 = 108. Sunil alone = 108 / 4 = 27 days.

Hard CAT 2023 · Slot 3

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

(1) 20
(2) 21
(3) 16
(4) 17

Show solution

(2) 21. Take total work = LCM(7, 15, 6) = 630 units. Combined efficiency of all three < 630/7 = 90/day. With 6 days all-three + 3 days Rakshita: 630 = (all-three)·6 + (Rakshita)·3 ⇒ Rakshita's efficiency > 630 − 90·2 = 30/day. So her solo time < 630/30 = 21 days. Hence she cannot take 21 days.

Hard CAT 2023 TITA

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

Show solution

36 days. 1/G + 1/S = 1/20. The make-up condition: 40% of G's rate = 50% of S's rate ⇒ (40/100)/G = (50/100)/S ⇒ 1/S = (4/5)(1/G). Substitute: 1/G + (4/5)(1/G) = 1/20 ⇒ (9/5)/G = 1/20 ⇒ G = 36. Gautam is the faster worker ⇒ 36 days.

CAT 2002 · Pipes & Cisterns

Moderate CAT 2002

Three small pumps and one large pump are filling a tank. Each of the three small pumps works at ⅔ʳᵈ the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?

(1) 4/7
(2) 1/3
(3) 2/3
(4) 3/4

Show solution

(2) 1/3. Let large pump = 3 units/hr ⇒ each small = 2 units/hr; three small = 6 units. All four together = 3 + 6 = 9 units/hr. Ratio of large-alone rate to all-four rate = 3/9 = 1/3. Since time is inversely proportional to rate, all four take 1/3 the time.

CAT 2017 · Pipes & Cisterns

Easy CAT 2017

A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

(1) 20
(2) 30
(3) 40
(4) 45

Show solution

(1) 20 hours. Let capacity = 40 L. Inlet = 40/8 = 5 L/hr; inlet + outlet = 40/10 = 4 L/hr ⇒ outlet drains 1 L/hr. Time to empty full tank = 40 hrs, so half-full = 40/2 = 20 hours.

CAT 2018 · Pipes & Cisterns

Hard CAT 2018 TITA

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?

Show solution

10 hours. Let filling = x, draining = −y. Capacity = (6x − 5y)·6 = (5x − 6y)·60 ⇒ 6x − 5y = 50x − 60y ⇒ 44x = 55y ⇒ x = 1.25y. Capacity = (6x − 5y)·6 = (7.5y − 5y)·6 = 15y. Net of 2 fill + 1 drain = 2x − y = 1.5y. Time = 15y / 1.5y = 10 hours.

Hard CAT 2018 TITA

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?

Show solution

48 minutes. Let A = a, B = b. Capacity = 0.5(10a + 45b) = (8a + 18b) ⇒ 10a + 45b = 16a + 36b ⇒ 6a = 9b. Capacity = 8a + 18b = 8a + 12a = 20a. Net of 7A + 27B = 7a + 27b = 7a + 18a = 25a. Time = 20a / 25a = 0.8 hr = 48 minutes.

CAT 2021 · Pipes & Cisterns

Hard CAT 2021 · Slot 2

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

(1) 264
(2) 140
(3) 120
(4) 144

Show solution

(4) 144 minutes. Let A fill the empty tank in x hours and B empty the full tank in y hours. Case I (2 p.m.→10 p.m., A runs 8 hrs, B runs 7 hrs): 8/x − 7/y = 1. Case II (2 p.m.→6 p.m., A runs 4 hrs, B runs 2 hrs): 4/x − 2/y = 1. Subtracting gives 4/x = 5/y. Substituting back: 8/x − 7/y = 1 with y = 5x/4 ⇒ x = 12/5 hr = 2 hr 24 min = 144 minutes (this is pipe A's solo fill time).

CAT 2023 · Pipes & Cisterns

Hard CAT 2023 · Slot 1

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 p.m. On Tuesday, B alone completed filling the tank at 6 p.m. On Wednesday, A alone worked till 5 p.m., and then B worked alone from 5 p.m. to 7 p.m., to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

(1) 4:48 p.m.
(2) 4:12 p.m.
(3) 4:24 p.m.
(4) 4:36 p.m.

Show solution

(3) 4:24 p.m. Let A alone take t hrs, B alone t − 2 hrs. From Wednesday's split: (t − 3)/t + 2/(t − 2) = 1 ⇒ t² − 3t + 6 = t² − 2t ⇒ t = 6. So the tank empties at 8 p.m. − 6 = 2 p.m. Both together: time = (6·4)/(6 + 4) = 2.4 hr = 2 h 24 min. Filled at 2 p.m. + 2:24 = 4:24 p.m.

Hard CAT 2023 · Slot 2

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

(1) 60
(2) 90
(3) 75
(4) 120

Show solution

(2) 90 minutes. Let A = x hrs, C = y hrs, B = (x − 1) hrs. Case 1: 1/x + 1/y − 1/(x−1) = 1/2. Case 2 (B 1 hr, C total 2¼ hrs): combining gives x(x − 1) = 18(x − 1) − 10x ⇒ x² − 9x + 18 = 0 ⇒ x = 3 or 6. Since A < 5 hrs, x = 3. Then 1/3 + 1/2 + 1/y = ... ⇒ y = 3/2 hr = 90 minutes.

CAT 2024 & 2025, recent

Fresh questions distributed from the real CAT 2024 & CAT 2025 papers into this chapter.

HardCAT 2025 · Slot 1

Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is

(A) 34400
(B) 38880
(C) 47040
(D) 38400

Show solution

(D) 38400. Tarun is the cheapest at Rs 2160/day with rate 1/15. Use Tarun for all 10 days: he does 10 × 1/15 = 2/3 of the task. The remaining 1/3 is done by the next-cheapest, Varun (rate 1/21): days = (1/3) ÷ (1/21) = 7 days. Minimum cost = 10 × 2160 + 7 × 2400 = 21600 + 16800 = Rs 38400.

HardCAT 2025 · Slot 2 TITA

Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is

Show solution

340. Let Chandan = 1, Bipin = 2, Ankita = 4 units/day. First 20 days all three: (4 + 2 + 1) × 20 = 140 units. Remaining 40 days Ankita + Chandan: (4 + 1) × 40 = 200 units. Total work = 140 + 200 = 340 units. Chandan alone = 340 ÷ 1 = 340 days.

HardCAT 2025 · Slot 3

Teams A, B, and C consist of five, eight, and ten members, respectively, such that every member within a team is equally productive. Working separately, teams A, B, and C can complete a certain job in 40 hours, 50 hours, and 4 hours, respectively. Two members from team A, three members from team B, and one member from team C together start the job, and the member from team C leaves after 23 hours. The number of additional member(s) from team B, that would be required to replace the member from team C, to finish the job in the next one hour, is

(A) 1
(B) 2
(C) 3
(D) 4

Show solution

(B) 2. Per-member rates: team A = (1/40)/5 = 1/200, team B = (1/50)/8 = 1/400, team C = (1/4)/10 = 1/40 per hour. First 23 hours, 2 of A + 3 of B + 1 of C work: per hour = 2/200 + 3/400 + 1/40 = 4/400 + 3/400 + 10/400 = 17/400. In 23 hours = 391/400, so remaining = 9/400. For the next 1 hour, C leaves; 2 of A + 3 of B already give 7/400, leaving 2/400 to be covered by extra B members at 1/400 each ⇒ 2 members.

ModerateCAT 2025 · Slot 3

The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to

(A) 78
(B) 73
(C) 71
(D) 76

Show solution

(B) 73. Let flow of A = 4x/hr. Pipe A fills in 15 hours ⇒ tank capacity = 15 × 4x = 60x. All three together = 4x + 9x + 36x = 49x/hr. Time = 60x ÷ 49x = 60/49 hr = (60/49) × 60 = 3600/49 ≈ 73.5 minutes, nearest to 73.