Katrina walks down an up-escalator and counts 150 steps. Priyanka walks up the same escalator and counts 75 steps. Ka...
In most of the questions on Time and Work, an initial scenario is given where the work done by a few people/groups/objects is compared. The question then goes on to describe another scenario, where the work done and/or time taken by one of the groups is given and the work done/time taken by another group is asked.
Let us take an example:
A can complete a project in 10 days. B can complete the same project in 15 days. Find the time taken to complete the project if A and B work on it together.
To solve this problem, the method generally used is:
Work done by A in one day = 1/10
Work done by B in one day = 1/15
Work done by A and B together in one day = 1/10 + 1/15 = 5/30 = 1/6
Therefore, they can complete the project in 1/(1/6) days = 6 days.
Now, let us take a slightly different approach to solve the problem:
A can complete the project in 10 days and B can complete the project in 15 days. The LCM of 10 and 15 is 30. So, think of the work as consisting of 30 units. Since A can complete it in 10 days, A's work per day = 3 units. Similarly, since B can complete the work in 15 days, B's work per day = 2 units. So, the work done by A and B together is 3+2 = 5 units. Since the work that needs to be done to complete the project is 30 units, the time taken by A and B together = 30/5 = 6 days.
What we have done here is convert a problem of fractions into a problem of whole numbers by multiplying each number with the LCM. The advantage with this slightly different method is that as the difficulty of the problem increases, the solution becomes that much easier.
Consider another example:
Three taps can fill a tank in 1 hour, 2 hours and 5 hours respectively. Another 3 taps can empty the tank in 6, 8 and 10 hours respectively. If all the taps are opened simultaneously, find the time in which the empty tank gets filled.
The traditional method involves dealing with 1, 1/2, 1/5 and fractions of the like. If we use the second method, we can solve the problem more easily:
The LCM of 1,2,5,6,8 and 10 is 120. So, let the total work done in filling the tap be 120 units. The work done by the taps in one hour are 120 units, 60 units, 24 units, -20 units, -15 units and -12 units respectively. The negative signs are for the taps that drain the tank. So, the work done by all the taps together is 120 + 60 + 24 - 20 - 15 - 12 = 157 units. Therefore, the time taken to fill the tank = 120/157 hours.
Problems involving comparison of work done by man, woman and child
Another type of question that is frequently asked is comparison of work between man, woman and child.
Consider the following example:
4 men or 8 women or 12 children can complete a project in 100 days. Find the time taken to complete the work if 6 men, 16 women and 12 children together work on the project.
Let the work be 2400 units. Therefore, work done by 4 men in one day = 2400/100 = 24 units, or work done by one man in one day = 6 units. Similarly, work done by one woman in one day = 3 units and work done by one child in one day = 2 units.
Therefore, work done by 6 men, 16 women and 12 children in one day = 36 + 48 + 24 = 108 units. Therefore, the time required to complete the project = 2400/108 = 200/9 days.
Consider a slight variation of the above problem:
A project can be completed by 4 men in 20 days or by 6 women in 25 days or by 24 children in 50 days. Find the time taken to complete the project if man, woman and child work on successive days i.e., one man works one the first day, one woman works on the second day, one child works on the third day and the cycle keeps repeating.
Let the number of units of work required be 6000. So, work done by 4 men = 300 units per day => word done by one man = 300/4 = 75 units per day. Similarly, work done by one woman = 6000/25/6 = 40 units per day and work done by one child = 6000/50/24 = 5 units per day. So, work done in one cycle (3 days) = 75 + 40 + 5 = 120 units. Therefore, number of cycles required = 6000/120 = 50 cycles. Number of days required = 50*3 = 150 days.
Substitution Problems
One variety of problems that can be tricky are substitution problems.
Consider the following example:
A man can complete a project in 100 days. A woman can complete it in 120 days. A group of 20 men start working on the project. After the first day, one man is replaced by a woman. On the second day, one more man is replaced by a woman. This process keeps repeating. Find the number of days required to complete the project.
Let the number of units of work involved be 2400. Work done by a man in one day = 24. Work done a woman in one day = 20. Since 20 men work on day 1, the work done on day 1 = 480. The next day, work done by the group = 19*24 + 20 = 476. On day 3, the work done = 472 and so on. In 5 days, the work done = 2360. Therefore, the work is completed on the 6th day.
Solved Examples
1) 10 men can complete a piece of work in 40 days. 15 women can complete the same work in 30 days. Find the time in which 5 men and 9 women, working together, will complete the same work.
Solution: Let the total work be 1200 units. Work done by 10 men in one day = 30 units. Work done by one man in one day = 3 units.
Work done by 15 women in one day = 40 units. Work done by one woman in one day = 2(2/3) units.
Work done by 5 men and 9 women in one day = 15+24 = 39 units. Therefore, total time needed to complete the work = 1200/39 = 30(30/39) days.
2) A tap takes 3 hours to fill a tank. Another tap takes 4 hours to empty the tank. If both the taps are opened simultaneously, find the time taken to fill a tank which is 40% full.
Solution: Let the work required be 12 units. Work done by first tap = 4 units/hour. Work done by second tap = -3units/hour. Therefore, work done by both the taps together = 1 unit/hour.
Work needed to be done = 60% of 12 units = 7.2 units. Therefore, time taken = 7.2/1 = 7 hours 12 minutes.
Maruti Konduri is an alumnus of IIT Bombay and IIM Ahmedabad and is the co-founder of the CAT preparation site Cracku (http://cracku.in)
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