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Arithmetic II

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Time & Work

Time And Work

MODULES

Basics & Worker-Day Method
Worker-Day Method
Unitary Method
Relative Efficiencies
Negative Work, Alternating Work & Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Time & Work 1
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Time & Work 2
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Time & Work 3
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PRACTICE

Time & Work : Level 1
Time & Work : Level 2
Time & Work : Level 3
ALL MODULES

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

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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 222 hours and 444 hours while drain C can empty a tank in 666 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 444 hours and tap Y can fill the same tank in 666 hours. A drain in the bottom of the tank can empty a full tank in 121212 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 111 hour are 14\dfrac{1}{4}41​ and 16\dfrac{1}{6}61​ respectively.

Portion of tank emptied by the drain in 111 hour is 112\dfrac{1}{12}121​.

Portion of tank filled in 111 hour =14+16−112=13= \dfrac{1}{4} + \dfrac{1}{6} - \dfrac{1}{12} = \dfrac{1}{3}=41​+61​−121​=31​

Time taken to fill the tank =3= 3=3 hours

Alternatively

Let the tank's capacity be LCM(4,6,12)=12(4, 6, 12) = 12(4,6,12)=12 units.

Taps X and Y fill the tank at the rate of 124\dfrac{12}{4}412​= 3 units/hr and 126\dfrac{12}{6}612​ = 2 units/hr respectively.

The drain empties the tank at the rate of 1212\dfrac{12}{12}1212​ = 1 unit/hr

When all are open, rate of filling =3+2−1= 3 + 2 - 1=3+2−1 = 4 units/hr

Time taken to fill the tank =124=3= \dfrac{12}{4} = 3=412​=3 hours

Answer: 333 hours


Example 20

Pipes A and B take 555 hours and 888 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 222 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 xxx hours to empty the tank.

Portion of the tank filled by the 222 pipes in 222 hours is 2(15+18)=13202\left( \dfrac{1}{5} + \dfrac{1}{8} \right) = \dfrac{13}{20}2(51​+81​)=2013​

As it takes 222 more hours to fill the remainder of the tank, which is 1−2640=7201 - \dfrac{26}{40} = \dfrac{7}{20}1−4026​=207​.

2(15+18−1x)=7202 \left( \dfrac{1}{5} + \dfrac{1}{8} - \dfrac{1}{x} \right) = \dfrac{7}{20} 2(51​+81​−x1​)=207​

⇒1340−740=1x \dfrac{13}{40} - \dfrac{7}{40} = \dfrac{1}{x}4013​−407​=x1​

⇒ x=203 x = \dfrac{20}{3}x=320​ = 6 hrs 40 mins

Alternatively

Let the tank's capacity be LCM(5,8)=40(5, 8) = 40(5,8)=40 units.

Pipes A and B fill the tank at the rate of 405\dfrac{40}{5}540​ = 8 units/hr and 408\dfrac{40}{8}840​ = 5 units/hr respectively.

Volume filled by A and B in 222 hours =2×(8+5)=26= 2 \times (8 + 5) = 26=2×(8+5)=26 units

Let the leak empty the tank at the rate of x\bm{x}x units/hr.
Volume of tank that is empty =40−26= 40 - 26=40−26 = 14 units

Volume filled by A, B and the leak in the next 222 hours =2×(8+5−x)=14= 2 \times (8 + 5 - x) = 14=2×(8+5−x)=14

⇒ 26−2x=14 26 - 2\text{x} = 1426−2x=14
⇒ x x x = 6 units/hr

Time taken by the leak to empty the tank =406=203= \dfrac{40}{6} = \dfrac{20}{3}=640​=320​ = 6 hrs 40 mins

Answer: 666 hrs 404040 mins


Example 21

A pipe usually takes 333 hours to fill an empty tank with water. However, on a particular day, it took 7.57.57.5 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 ttt.

13−1t=17.5\dfrac{1}{3} - \dfrac{1}{t} = \dfrac{1}{7.5}31​−t1​=7.51​ ⇒ 1t=13−215\dfrac{1}{t} = \dfrac{1}{3} - \dfrac{2}{15}t1​=31​−152​

⇒t=15 t = 15t=15

Answer: 555

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 12,1512, 1512,15 and 242424 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 333 days. In how many days is the work completed?

Solution

Portion of work completed every 333 days =112+115+124=10+8+5120=23120= \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{24} = \dfrac{10 + 8 + 5}{120} = \dfrac{23}{120}=121​+151​+241​=12010+8+5​=12023​

Let's look at the work as involving 120120120 parts. A, B and C can finish 10,810, 810,8 and 555 parts in a day respectively and in 333 days 232323 parts of work get completed.

In 5 such cycles involving 151515 days, 5×23=1155 \times 23 = 1155×23=115 parts get completed and 555 are remaining.

It's A's turn now, who can complete 101010 parts in a day. Since there are 555 parts left, A can complete in 510=0.5\dfrac{5}{10} = 0.5105​=0.5 day. Therefore, the work gets completed in 15.515.515.5 days.

Answer: 15.515.515.5


Example 23

A and C are pipes that take 101010 minutes and 606060 minutes to fill a tank. B and D are leaks that take 404040 and 242424 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 444 minutes, portion of tank filled =110−140+160−124=12−3+2−5120=6120= \dfrac{1}{10} - \dfrac{1}{40} + \dfrac{1}{60} - \dfrac{1}{24} = \dfrac{12 - 3 + 2 - 5}{120} = \dfrac{6}{120}=101​−401​+601​−241​=12012−3+2−5​=1206​

Once again, the denominator can be looked at as the total work (i.e., 120 parts). In one cycle of 444 minutes, 666 parts are filled. Parts of tank filled by Pipe A, Leak B, Pipe C and Leak D are +12,−3,+2+12, -3, +2+12,−3,+2 and −5-5−5 respectively.

The maximum that a pipe can fill in 111 minute is Pipe A which is 121212 parts. So, we can find out the number of cycles that fill 120−12=108120 - 12 = 108120−12=108 parts to provide for the possibility of a pipe filling the rest of the tank.

Therefore, in 181818 cycles or 727272 minutes, 18×6=10818 \times 6 = 10818×6=108 parts are filled and 121212 parts are remaining. As it is A's turn for the 73rd73^\text{rd}73rd minute and it fills 121212 parts, the work is complete in 737373 minutes.

Answer: 737373 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]

202020 minutes
303030 minutes
404040 minutes
505050 minutes

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

Let the time taken by all 333 together to complete the work be ttt.
Time taken by Asit, Arnold and Afzal are t+6,t+1t + 6, t + 1t+6,t+1 and 2t2t2t respectively.

When all 333 work together,
1t+6+1t+1+12t=1t\dfrac{1}{t + 6} + \dfrac{1}{t + 1} + \dfrac{1}{2t} = \dfrac{1}{t}t+61​+t+11​+2t1​=t1​

⇒ 1t+6+1t+1=1t−12t \dfrac{1}{t + 6} + \dfrac{1}{t + 1} = \dfrac{1}{t} - \dfrac{1}{2t}t+61​+t+11​=t1​−2t1​

t+1+6(t+6)(t+1)=12t {t + 1 + 6}{(t + 6)(t + 1)} = \dfrac{1}{2t}t+1+6(t+6)(t+1)=2t1​
⇒ 4t2+14t=t2+7t+6 4t^{2} + 14t = t^{2} + 7{t} + {6}4t2+14t=t2+7t+6

⇒ 3t2+7t−6=0 3t^{2} + 7t - 6 = 03t2+7t−6=0

⇒ 3t2+9t−2t−6=0 3t^{2} + 9t - 2t - 6 = 03t2+9t−2t−6=0

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

t=23,−3 t = \dfrac{2}{3}, -3t=32​,−3 [Negative value is rejected]

Therefore, time taken together is 23\dfrac{2}{3}32​ hours or 404040 minutes.

Answer: (333) 404040 minutes

Question 2

A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 888 hours while an outlet pipe can empty the full tank in 121212 hours. If all pipes are left open simultaneously, it takes 666 hours to fill the empty tank. What is the relationship between M and N?
[XAT 2016]

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

Observation/Strategy
111) 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
M8−N12=16 \dfrac{M}{8} - \dfrac{N}{12} = \dfrac{1}{6}8M​−12N​=61​

⇒ 3M−2N24=16 \dfrac{3M - 2N}{24} = \dfrac{1}{6}243M−2N​=61​

3M−2N=43M - 2N = 43M−2N=4

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

Answer: (555) None of the above

Note that (M, N) can take values like (2,12, 12,1), (4,44, 44,4), etc. The ratios for these are 2:1,1:12 : 1, 1 : 12:1,1:1, 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 303030 minutes for all three of them cooking together to complete 505050 percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after 111 hour and the younger son cooks for further 333 hours. If the mother needs 111 hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?
[IIFT 2013]

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

Observation/Strategy
1) As 333 of them complete 50%50 \%50% of the work in 303030 mins, they complete the full work in 111 hour.
2) Work gets completed if elder son works for 111 hour and younger son works for 444 hours.
3) Mother takes 111 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 xxx and yyy hours respectively. Therefore, the mother takes x−1x - 1x−1 hours to complete the work.

When the elder and younger sons work for 111 and 444 hours respectively,
1x+4y=11y=14−14x\dfrac{1}{x} + \dfrac{4}{y} = 1 \dfrac{1}{y} = \dfrac{1}{4} - \dfrac{1}{4x}x1​+y4​=1y1​=41​−4x1​

When all three of them work together,
1x−1+1x+1y=11x−1+1x+14−14x=1\dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{y} = 1 \dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{4} - \dfrac{1}{4x} = 1x−11​+x1​+y1​=1x−11​+x1​+41​−4x1​=1

1x−1+34x=34(4x+3x−34(x2−x)=34 \dfrac{1}{x - 1} + \dfrac{3}{4x} = \dfrac{3}{4} ( \dfrac{4x + 3x - 3}{4(x^{2} - x}) = \dfrac{3}{4}x−11​+4x3​=43​(4(x2−x4x+3x−3​)=43​

⇒ 7x−3=3x2−3x3x2−10x+3=0 7x - 3 = 3x^{2} - 3x 3x^{2} - 10x + 3 = 07x−3=3x2−3x3x2−10x+3=0

(3x−1)(x−3)=0 (3x - 1)(x - 3) = 0(3x−1)(x−3)=0
x =13,3= \dfrac{1}{3}, 3=31​,3 [13\dfrac{1}{3}31​ is not possible as this would result in time taken by mother to be negative]

Time taken by mother working alone =3−1=2= 3 - 1 = 2=3−1=2 hours

Portion of work completed by mother in 111 hour =12=50%= \dfrac{1}{2} = 50 \%=21​=50%

Answer: (2) 50%50 \%50%

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