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Arithmetic II

Time & Work

ALL MODULES

CAT 2025 Lesson : Time & Work - Negative Work, Alternating Work & Past Questions

5. Negative rate of completion

Negative rate of completion questions are typically pipes, drains/leaks and tanks based. For instance, Pipe A and Pipe B can fill a tank in

2

hours and

4

hours while drain C can empty a tank in

6

hours. If all three are open at the same time, how long does it take to fill an empty tank?

In such questions the efficiency of a leak, or something undoing the work, will be negative.

Example 19

Tap X can fill a tank in

4

hours and tap Y can fill the same tank in

6

hours. A drain in the bottom of the tank can empty a full tank in

12

hours. If the two taps and the drain are opened at the same time, how long does it take an empty tank to be filled?

Solution

Portion of tank filled by taps X and Y in

1

hour are

\dfrac{1}{4}

and

\dfrac{1}{6}

respectively.

Portion of tank emptied by the drain in

1

hour is

\dfrac{1}{12}

.

Portion of tank filled in

1

hour

= \dfrac{1}{4} + \dfrac{1}{6} - \dfrac{1}{12} = \dfrac{1}{3}

Time taken to fill the tank

= 3

hours

Alternatively

Let the tank's capacity be LCM

(4, 6, 12) = 12

units.

Taps X and Y fill the tank at the rate of

\dfrac{12}{4}

= 3 units/hr and

\dfrac{12}{6}

= 2 units/hr respectively.

The drain empties the tank at the rate of

\dfrac{12}{12}

= 1 unit/hr

When all are open, rate of filling

= 3 + 2 - 1

= 4 units/hr

Time taken to fill the tank

= \dfrac{12}{4} = 3

hours

Answer:

3

hours

Example 20

Pipes A and B take

5

hours and

8

hours respectively to fill an empty tank. Both the pipes are turned on at the same time. Exactly after two hours, a leak develops due to the pressure in the tank. It, therefore, takes

2

more hours for the tank to be filled. If the water flowing out of the leak was at a steady rate, how long would it take a filled tank to be emptied by the leak?

Solution

Let the leak take

x

hours to empty the tank.

Portion of the tank filled by the

2

pipes in

2

hours is

2\left( \dfrac{1}{5} + \dfrac{1}{8} \right) = \dfrac{13}{20}

As it takes

2

more hours to fill the remainder of the tank, which is

1 - \dfrac{26}{40} = \dfrac{7}{20}

2 \left( \dfrac{1}{5} + \dfrac{1}{8} - \dfrac{1}{x} \right) = \dfrac{7}{20}

⇒

\dfrac{13}{40} - \dfrac{7}{40} = \dfrac{1}{x}

⇒

x = \dfrac{20}{3}

= 6 hrs 40 mins

Alternatively

Let the tank's capacity be LCM

(5, 8) = 40

units.

Pipes A and B fill the tank at the rate of

\dfrac{40}{5}

= 8 units/hr and

\dfrac{40}{8}

= 5 units/hr respectively.

Volume filled by A and B in

2

hours

= 2 \times (8 + 5) = 26

units

Let the leak empty the tank at the rate of

\bm{x}

units/hr.
Volume of tank that is empty

= 40 - 26

= 14 units

Volume filled by A, B and the leak in the next

2

hours

= 2 \times (8 + 5 - x) = 14

⇒

26 - 2\text{x} = 14

⇒

x

= 6 units/hr

Time taken by the leak to empty the tank

= \dfrac{40}{6} = \dfrac{20}{3}

= 6 hrs 40 mins

Answer:

6

hrs

40

mins

Example 21

A pipe usually takes

3

hours to fill an empty tank with water. However, on a particular day, it took

7.5

hours to fill the tank due to a leak it had developed. Then, how many hours would the leak take to empty a full tank?

Solution

Let the time taken by the leak to empty a full tank be

t

\dfrac{1}{3} - \dfrac{1}{t} = \dfrac{1}{7.5}

⇒

\dfrac{1}{t} = \dfrac{1}{3} - \dfrac{2}{15}

⇒

t = 15

Answer:

5

6. People or objects working alternately

These questions involve objects or group of objects with different efficiencies taking turns and working. In these questions you need to calculate the portion of work completed in a cycle and then find out the point where a single person or object's work would render the work complete. The Parts Method is best to use here.

Example 22

When working alone, A, B and C take

12, 15

and

24

days to complete a project. A works alone on the project on the first day, B works alone on the second day and C works alone on the third day. This pattern continues with each of the individuals working alone every

3

days. In how many days is the work completed?

Solution

Portion of work completed every

3

days

= \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{24} = \dfrac{10 + 8 + 5}{120} = \dfrac{23}{120}

Let's look at the work as involving

120

parts. A, B and C can finish

10, 8

and

5

parts in a day respectively and in

3

days

23

parts of work get completed.

In 5 such cycles involving

15

days,

5 \times 23 = 115

parts get completed and

5

are remaining.

It's A's turn now, who can complete

10

parts in a day. Since there are

5

parts left, A can complete in

\dfrac{5}{10} = 0.5

day. Therefore, the work gets completed in

15.5

days.

Answer:

15.5

Example 23

A and C are pipes that take

10

minutes and

60

minutes to fill a tank. B and D are leaks that take

40

and

24

minutes to empty a tank. At any given time only a tap or a drain is open. A is turned on for the first minute, B is then turned on for the second minute, C for the third, D for the fourth, A for fifth, B for the sixth and so on. This is repeated till the tank is full. How long does it take for the tank to be filled?

Solution

In one cycle of

4

minutes, portion of tank filled

= \dfrac{1}{10} - \dfrac{1}{40} + \dfrac{1}{60} - \dfrac{1}{24} = \dfrac{12 - 3 + 2 - 5}{120} = \dfrac{6}{120}

Once again, the denominator can be looked at as the total work (i.e., 120 parts). In one cycle of

4

minutes,

6

parts are filled. Parts of tank filled by Pipe A, Leak B, Pipe C and Leak D are

+12, -3, +2

and

-5

respectively.

The maximum that a pipe can fill in

1

minute is Pipe A which is

12

parts. So, we can find out the number of cycles that fill

120 - 12 = 108

parts to provide for the possibility of a pipe filling the rest of the tank.

Therefore, in

18

cycles or

72

minutes,

18 \times 6 = 108

parts are filled and

12

parts are remaining. As it is A's turn for the

73^\text{rd}

minute and it fills

12

parts, the work is complete in

73

minutes.

Answer:

73

minutes

7. Past Questions

Question 1

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?
[CAT 2001]

		$20$ minutes
		$30$ minutes
		$40$ minutes
		$50$ minutes

Observation/Strategy

1

) We can represent the individual time taken in terms of the total time taken and form an equation.

Let the time taken by all

3

together to complete the work be

t

.
Time taken by Asit, Arnold and Afzal are

t + 6, t + 1

and

2t

respectively.

When all

3

work together,

\dfrac{1}{t + 6} + \dfrac{1}{t + 1} + \dfrac{1}{2t} = \dfrac{1}{t}

⇒

\dfrac{1}{t + 6} + \dfrac{1}{t + 1} = \dfrac{1}{t} - \dfrac{1}{2t}

{t + 1 + 6}{(t + 6)(t + 1)} = \dfrac{1}{2t}

⇒

4t^{2} + 14t = t^{2} + 7{t} + {6}

⇒

3t^{2} + 7t - 6 = 0

⇒

3t^{2} + 9t - 2t - 6 = 0

⇒

(3t - 2)(t + 3) = 0

t = \dfrac{2}{3}, -3

[Negative value is rejected]

Therefore, time taken together is

\dfrac{2}{3}

hours or

40

minutes.

Answer: (

3

)

40

minutes

Question 2

A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in

8

hours while an outlet pipe can empty the full tank in

12

hours. If all pipes are left open simultaneously, it takes

6

hours to fill the empty tank. What is the relationship between M and N?
[XAT 2016]

		M : N $= 1 : 1$
		M : N $= 2 : 1$
		M : N $= 2 : 3$
		M : N $= 3 : 2$
		None of the above

Observation/Strategy

1

) From the given data, we will be able to form just 1 equation on portion of work completed in an hour. We shall try finding the ratio with this.

Portion of tank filled in an hour when M inlet and N outlet pipes are opened is

\dfrac{M}{8} - \dfrac{N}{12} = \dfrac{1}{6}

⇒

\dfrac{3M - 2N}{24} = \dfrac{1}{6}

3M - 2N = 4

As there is a constant in this equation, we will not be able to find M : N.

Answer: (

5

) None of the above

Note that (M, N) can take values like (

2, 1

), (

4, 4

), etc. The ratios for these are

2 : 1, 1 : 1

, etc. So, there is no unique ratio.

Question 3

A mother along with her two sons is entrusted with the task of cooking Biryani for a family get together. It takes

30

minutes for all three of them cooking together to complete

50

percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after

1

hour and the younger son cooks for further

3

hours. If the mother needs

1

hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?
[IIFT 2013]

		$33.33 \%$
		$50 \%$
		$66.67 \%$
		None of the above

Observation/Strategy
1) As

3

of them complete

50 \%

of the work in

30

mins, they complete the full work in

1

hour.
2) Work gets completed if elder son works for

1

hour and younger son works for

4

hours.
3) Mother takes

1

hour less than elder son to complete the work.

Let the time taken by elder son and younger son to individually complete the work be

x

and

y

hours respectively. Therefore, the mother takes

x - 1

hours to complete the work.

When the elder and younger sons work for

1

and

4

hours respectively,

\dfrac{1}{x} + \dfrac{4}{y} = 1 \dfrac{1}{y} = \dfrac{1}{4} - \dfrac{1}{4x}

When all three of them work together,

\dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{y} = 1 \dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{4} - \dfrac{1}{4x} = 1

\dfrac{1}{x - 1} + \dfrac{3}{4x} = \dfrac{3}{4} ( \dfrac{4x + 3x - 3}{4(x^{2} - x}) = \dfrac{3}{4}

⇒

7x - 3 = 3x^{2} - 3x 3x^{2} - 10x + 3 = 0

(3x - 1)(x - 3) = 0

= \dfrac{1}{3}, 3

[

\dfrac{1}{3}

is not possible as this would result in time taken by mother to be negative]

Time taken by mother working alone

= 3 - 1 = 2

hours

Portion of work completed by mother in

1

hour

= \dfrac{1}{2} = 50 \%

Answer: (2)

50 \%

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Quant

Arithmetic II

Time & Work

ALL MODULES

CAT 2025 Lesson : Time & Work - Negative Work, Alternating Work & Past Questions

5. Negative rate of completion

Negative rate of completion questions are typically pipes, drains/leaks and tanks based. For instance, Pipe A and Pipe B can fill a tank in

2

hours and

4

hours while drain C can empty a tank in

6

hours. If all three are open at the same time, how long does it take to fill an empty tank?

In such questions the efficiency of a leak, or something undoing the work, will be negative.

Example 19

Tap X can fill a tank in

4

hours and tap Y can fill the same tank in

6

hours. A drain in the bottom of the tank can empty a full tank in

12

hours. If the two taps and the drain are opened at the same time, how long does it take an empty tank to be filled?

Solution

Portion of tank filled by taps X and Y in

1

hour are

\dfrac{1}{4}

and

\dfrac{1}{6}

respectively.

Portion of tank emptied by the drain in

1

hour is

\dfrac{1}{12}

.

Portion of tank filled in

1

hour

= \dfrac{1}{4} + \dfrac{1}{6} - \dfrac{1}{12} = \dfrac{1}{3}

Time taken to fill the tank

= 3

hours

Alternatively

Let the tank's capacity be LCM

(4, 6, 12) = 12

units.

Taps X and Y fill the tank at the rate of

\dfrac{12}{4}

= 3 units/hr and

\dfrac{12}{6}

= 2 units/hr respectively.

The drain empties the tank at the rate of

\dfrac{12}{12}

= 1 unit/hr

When all are open, rate of filling

= 3 + 2 - 1

= 4 units/hr

Time taken to fill the tank

= \dfrac{12}{4} = 3

hours

Answer:

3

hours

Example 20

Pipes A and B take

5

hours and

8

hours respectively to fill an empty tank. Both the pipes are turned on at the same time. Exactly after two hours, a leak develops due to the pressure in the tank. It, therefore, takes

2

more hours for the tank to be filled. If the water flowing out of the leak was at a steady rate, how long would it take a filled tank to be emptied by the leak?

Solution

Let the leak take

x

hours to empty the tank.

Portion of the tank filled by the

2

pipes in

2

hours is

2\left( \dfrac{1}{5} + \dfrac{1}{8} \right) = \dfrac{13}{20}

As it takes

2

more hours to fill the remainder of the tank, which is

1 - \dfrac{26}{40} = \dfrac{7}{20}

2 \left( \dfrac{1}{5} + \dfrac{1}{8} - \dfrac{1}{x} \right) = \dfrac{7}{20}

⇒

\dfrac{13}{40} - \dfrac{7}{40} = \dfrac{1}{x}

⇒

x = \dfrac{20}{3}

= 6 hrs 40 mins

Alternatively

Let the tank's capacity be LCM

(5, 8) = 40

units.

Pipes A and B fill the tank at the rate of

\dfrac{40}{5}

= 8 units/hr and

\dfrac{40}{8}

= 5 units/hr respectively.

Volume filled by A and B in

2

hours

= 2 \times (8 + 5) = 26

units

Let the leak empty the tank at the rate of

\bm{x}

units/hr.
Volume of tank that is empty

= 40 - 26

= 14 units

Volume filled by A, B and the leak in the next

2

hours

= 2 \times (8 + 5 - x) = 14

⇒

26 - 2\text{x} = 14

⇒

x

= 6 units/hr

Time taken by the leak to empty the tank

= \dfrac{40}{6} = \dfrac{20}{3}

= 6 hrs 40 mins

Answer:

6

hrs

40

mins

Example 21

A pipe usually takes

3

hours to fill an empty tank with water. However, on a particular day, it took

7.5

hours to fill the tank due to a leak it had developed. Then, how many hours would the leak take to empty a full tank?

Solution

Let the time taken by the leak to empty a full tank be

t

\dfrac{1}{3} - \dfrac{1}{t} = \dfrac{1}{7.5}

⇒

\dfrac{1}{t} = \dfrac{1}{3} - \dfrac{2}{15}

⇒

t = 15

Answer:

5

6. People or objects working alternately

Example 22

When working alone, A, B and C take

12, 15

and

24

3

days. In how many days is the work completed?

Solution

Portion of work completed every

3

days

= \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{24} = \dfrac{10 + 8 + 5}{120} = \dfrac{23}{120}

Let's look at the work as involving

120

parts. A, B and C can finish

10, 8

and

5

parts in a day respectively and in

3

days

23

parts of work get completed.

In 5 such cycles involving

15

days,

5 \times 23 = 115

parts get completed and

5

are remaining.

It's A's turn now, who can complete

10

parts in a day. Since there are

5

parts left, A can complete in

\dfrac{5}{10} = 0.5

day. Therefore, the work gets completed in

15.5

days.

Answer:

15.5

Example 23

A and C are pipes that take

10

minutes and

60

minutes to fill a tank. B and D are leaks that take

40

and

24

Solution

In one cycle of

4

minutes, portion of tank filled

= \dfrac{1}{10} - \dfrac{1}{40} + \dfrac{1}{60} - \dfrac{1}{24} = \dfrac{12 - 3 + 2 - 5}{120} = \dfrac{6}{120}

Once again, the denominator can be looked at as the total work (i.e., 120 parts). In one cycle of

4

minutes,

6

parts are filled. Parts of tank filled by Pipe A, Leak B, Pipe C and Leak D are

+12, -3, +2

and

-5

respectively.

The maximum that a pipe can fill in

1

minute is Pipe A which is

12

parts. So, we can find out the number of cycles that fill

120 - 12 = 108

parts to provide for the possibility of a pipe filling the rest of the tank.

Therefore, in

18

cycles or

72

minutes,

18 \times 6 = 108

parts are filled and

12

parts are remaining. As it is A's turn for the

73^\text{rd}

minute and it fills

12

parts, the work is complete in

73

minutes.

Answer:

73

minutes

7. Past Questions

Question 1

		$20$ minutes
		$30$ minutes
		$40$ minutes
		$50$ minutes

Observation/Strategy

1

) We can represent the individual time taken in terms of the total time taken and form an equation.

Let the time taken by all

3

together to complete the work be

t

.
Time taken by Asit, Arnold and Afzal are

t + 6, t + 1

and

2t

respectively.

When all

3

work together,

\dfrac{1}{t + 6} + \dfrac{1}{t + 1} + \dfrac{1}{2t} = \dfrac{1}{t}

⇒

\dfrac{1}{t + 6} + \dfrac{1}{t + 1} = \dfrac{1}{t} - \dfrac{1}{2t}

{t + 1 + 6}{(t + 6)(t + 1)} = \dfrac{1}{2t}

⇒

4t^{2} + 14t = t^{2} + 7{t} + {6}

⇒

3t^{2} + 7t - 6 = 0

⇒

3t^{2} + 9t - 2t - 6 = 0

⇒

(3t - 2)(t + 3) = 0

t = \dfrac{2}{3}, -3

[Negative value is rejected]

Therefore, time taken together is

\dfrac{2}{3}

hours or

40

minutes.

Answer: (

3

)

40

minutes

Question 2

A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in

8

hours while an outlet pipe can empty the full tank in

12

hours. If all pipes are left open simultaneously, it takes

6

hours to fill the empty tank. What is the relationship between M and N?
[XAT 2016]

		M : N $= 1 : 1$
		M : N $= 2 : 1$
		M : N $= 2 : 3$
		M : N $= 3 : 2$
		None of the above

Observation/Strategy

1

\dfrac{M}{8} - \dfrac{N}{12} = \dfrac{1}{6}

⇒

\dfrac{3M - 2N}{24} = \dfrac{1}{6}

3M - 2N = 4

As there is a constant in this equation, we will not be able to find M : N.

Answer: (

5

) None of the above

Note that (M, N) can take values like (

2, 1

), (

4, 4

), etc. The ratios for these are

2 : 1, 1 : 1

, etc. So, there is no unique ratio.

Question 3

A mother along with her two sons is entrusted with the task of cooking Biryani for a family get together. It takes

30

minutes for all three of them cooking together to complete

50

percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after

1

hour and the younger son cooks for further

3

hours. If the mother needs

1

hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?
[IIFT 2013]

		$33.33 \%$
		$50 \%$
		$66.67 \%$
		None of the above

Observation/Strategy
1) As

3

of them complete

50 \%

of the work in

30

mins, they complete the full work in

1

hour.
2) Work gets completed if elder son works for

1

hour and younger son works for

4

hours.
3) Mother takes

1

hour less than elder son to complete the work.

Let the time taken by elder son and younger son to individually complete the work be

x

and

y

hours respectively. Therefore, the mother takes

x - 1

hours to complete the work.

When the elder and younger sons work for

1

and

4

hours respectively,

\dfrac{1}{x} + \dfrac{4}{y} = 1 \dfrac{1}{y} = \dfrac{1}{4} - \dfrac{1}{4x}

When all three of them work together,

\dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{y} = 1 \dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{4} - \dfrac{1}{4x} = 1

\dfrac{1}{x - 1} + \dfrac{3}{4x} = \dfrac{3}{4} ( \dfrac{4x + 3x - 3}{4(x^{2} - x}) = \dfrac{3}{4}

⇒

7x - 3 = 3x^{2} - 3x 3x^{2} - 10x + 3 = 0

(3x - 1)(x - 3) = 0

= \dfrac{1}{3}, 3

[

\dfrac{1}{3}

is not possible as this would result in time taken by mother to be negative]

Time taken by mother working alone

= 3 - 1 = 2

hours

Portion of work completed by mother in

1

hour

= \dfrac{1}{2} = 50 \%

Answer: (2)

50 \%

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