Time and work shortcuts play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

1. If a person can do a piece of work in ‘m’ days, he can do (1/m) part of the work in 1 day. |

2. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio. |

3. If X is twice as good a workman as Y, then X will take half the time taken by Y to do a certain piece of work. |

4. Time and work are always in direct proportion. |

5. If two taps or pipes P and Q take ‘m’ and ‘n’ hours respectively to fill a cistern or tank, then the two pipes together fill (1/m)+(1/n) part of the tank in one hour. Entire tank is filled in mn/(m+n) hours. |

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic time and work problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve time and work problems. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to
learn few basic operations in this topic time and work and some additional tricks. Already we
are much clear with the four basic operations which we often use in math. They
are addition, subtraction, multiplication and division. Even though we are much
clear with these four basic operations, we have to be knowing some more stuff
to do the problems which are being asked from this topic in competitive exams.
The stuff which I have mentioned above is nothing but the tricks and shortcuts
which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve problems related to time and work. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as time and work shortcuts.

**Here, we are going to have some problems in which time and work shortcuts are applied. You can check your answer online and see step by step solution.**

1. A and B can complete a work in 12 days.**B and C can complete the same
work in 18 days. C and A can complete in 24 days. How many days will
take for A, B and C combined together to complete the same amount of
work?
**

From the given information, we can have

(A+B) can complete(1/12) part of the work in 1 day

(B+C) can complete (1/18) part of the work in 1 day

(A+C) can complete (1/24) part of the work in 1 day

By adding the three equations, we get,

(A+B)+(B+C)+(A+C)= 1/12+1/18+1/24

2A+2B+2C=(6+4+3)/72

2(A+B+C)=13/72

(A+B+C) can complete 13/144 part of the work in 1 day

Therefore (A+B+C) can together complete the work in 144/13 days

That is 11(1/3) days

(A+B) can complete(1/12) part of the work in 1 day

(B+C) can complete (1/18) part of the work in 1 day

(A+C) can complete (1/24) part of the work in 1 day

By adding the three equations, we get,

(A+B)+(B+C)+(A+C)= 1/12+1/18+1/24

2A+2B+2C=(6+4+3)/72

2(A+B+C)=13/72

(A+B+C) can complete 13/144 part of the work in 1 day

Therefore (A+B+C) can together complete the work in 144/13 days

That is 11(1/3) days

2. A & B can do a work in 15 days, B & C in 30 days and A & C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work?**
**

We can apply L.C.M method to solve this problem

Total work = 90 units (L.C.M of 15,30,18,9)

(A+B) can complete 6 units/day (90/15 = 6)

(B+C) can complete 3 units/day (90/30 = 3)

(A+C) can complete 5 units/day (90/18 = 5)

By adding, we get 2(A+B+C) = 14 units/day

(A+B+C) = 14/2 = 7 units/day

work done by A = (A+B+C)-(B+C)=7-3=4 units/day

work done by B = (A+B+C)-(A+C)=7-5=2 units/day

work done by C = (A+B+C)-(A+B)=7-6=1 unit/day

work done by (A+B+C)in 9 days = 9X7 = 63 units.

Balance work = 90-63 = 27 units

This 27 units of work to be completed B & C. Because A left after 9 days of work.

No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day)

Total work = 90 units (L.C.M of 15,30,18,9)

(A+B) can complete 6 units/day (90/15 = 6)

(B+C) can complete 3 units/day (90/30 = 3)

(A+C) can complete 5 units/day (90/18 = 5)

By adding, we get 2(A+B+C) = 14 units/day

(A+B+C) = 14/2 = 7 units/day

work done by A = (A+B+C)-(B+C)=7-3=4 units/day

work done by B = (A+B+C)-(A+C)=7-5=2 units/day

work done by C = (A+B+C)-(A+B)=7-6=1 unit/day

work done by (A+B+C)in 9 days = 9X7 = 63 units.

Balance work = 90-63 = 27 units

This 27 units of work to be completed B & C. Because A left after 9 days of work.

No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day)

3. A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

From the question, we have to consider an important thing.

That is, pipe B is faster than pipe A.

When two pipes are opened together, the tank will emptied.

So the right choice would be (A) or (B)

Total capacity of the tank = 60 units. (L.C.M of 10,6)

The tank is already two-fifth full.

That is,quantity of water is in the tank =(2/5)X60 = 24 units

If both the pipes are opened together, this 24 units will be emptied.

work done by pipe A = 60/10 = 6 units/min

work done by pipe B = 60/6 = -12 units/min (emptying the tank)

Adding the above two equations, we get(A+B)=-4 units/min

That is 4 units will be emptied per minute when both the pipes are opened together

Time taken to empty 24 units (2/5 of the tank)= 24/4=6 min

Time taken to empty the tank is 6 min. Option (A) is correct.

That is, pipe B is faster than pipe A.

When two pipes are opened together, the tank will emptied.

So the right choice would be (A) or (B)

Total capacity of the tank = 60 units. (L.C.M of 10,6)

The tank is already two-fifth full.

That is,quantity of water is in the tank =(2/5)X60 = 24 units

If both the pipes are opened together, this 24 units will be emptied.

work done by pipe A = 60/10 = 6 units/min

work done by pipe B = 60/6 = -12 units/min (emptying the tank)

Adding the above two equations, we get(A+B)=-4 units/min

That is 4 units will be emptied per minute when both the pipes are opened together

Time taken to empty 24 units (2/5 of the tank)= 24/4=6 min

Time taken to empty the tank is 6 min. Option (A) is correct.

4. A contractor decided to complete the work in 90 days and employed 50 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed the additional men,how many extra days would he have needed to complete the work?

Given Information:

The work has to completed in 90 days (as per schedule)

Total no. of men appointed initially = 50

50 men worked 20 days and completed a part of the work

The remaining work is completed by 70 men (50+20=70) in 70 days (90-20=70)

If the remaining work is completed by 50 men,

no. of days taken by them = (70X70)/50 = 98 days.

Hence, extra days needed = 98-70 = 28 days.

The work has to completed in 90 days (as per schedule)

Total no. of men appointed initially = 50

50 men worked 20 days and completed a part of the work

The remaining work is completed by 70 men (50+20=70) in 70 days (90-20=70)

If the remaining work is completed by 50 men,

no. of days taken by them = (70X70)/50 = 98 days.

Hence, extra days needed = 98-70 = 28 days.

5.Three taps A, B and C can fill a tank in 10, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

Total work = 60 units. (L.C.M of 10,15,20)

work done by the pipe A = 60/10 = 6 units/hr

work done by the pipe B = 60/15 = 4 units/hr

work done by the pipe C = 60/20 = 3 units/hr

(Given:A is open all the time,B and C are alternately)

1st hour: (A+B) = 10 units/hr

2nd hour: (A+C) = 9 units/hr

3rd hour: (A+B) = 10 units/hr

4th hour: (A+C) = 9 units/hr

5th hour: (A+B) = 10 units/hr

6th hour: (A+C) = 9 units/hr

When we add the above units, we get the total 57 units.

Apart from the 6 hours of operation, to get the total work 60 units, A has to work for half an hour

Because in one of hour work of A, we will get 6 units)

Hence, time taken to fill the tank = 6.5 hours.

work done by the pipe A = 60/10 = 6 units/hr

work done by the pipe B = 60/15 = 4 units/hr

work done by the pipe C = 60/20 = 3 units/hr

(Given:A is open all the time,B and C are alternately)

1st hour: (A+B) = 10 units/hr

2nd hour: (A+C) = 9 units/hr

3rd hour: (A+B) = 10 units/hr

4th hour: (A+C) = 9 units/hr

5th hour: (A+B) = 10 units/hr

6th hour: (A+C) = 9 units/hr

When we add the above units, we get the total 57 units.

Apart from the 6 hours of operation, to get the total work 60 units, A has to work for half an hour

Because in one of hour work of A, we will get 6 units)

Hence, time taken to fill the tank = 6.5 hours.

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