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Man Hour Work
Most of the questions asked in CAT from the chapter MAN HOUR WORK are complicated in nature and requires a good reasoning skill. First try to understand the basics and solve the questions given here as example before moving to the questions in Question Bank.
Let‘s take an example of building a bridge.
You have been told that 10 men worked for 10 days and 8 hours daily to complete the above task.
Now lets analyze few different cases by changing some of the above variables associated with the task.
Case1: You are asked to build a similar bridge at some other place but with less number of people (say 8) then it’s quite obvious that the number of days required shall increase.
(Note: Here we are assuming that each man does same amount of work in a given time frame)
Case 2: You have to build one more bridge with 10 men but the number of working hours reduced to 6 hr daily. In this case the number of days‘ll also increase.
Case 3: Let’s consider another case where we are in a hurry and required to finish the work in less number of days (say 5).
Then what we‘ll have to do?
We have to increase either the number of workers or the number of working hours per day or both.
Case 4: Finally let’s consider a situation where we are required to build more than one bridge (say 2).
Here if we want to build the bridge in 10 days then we would be required to increase either the number of workers or the number of working hours per day or both.
If we don’t increase the number of workers or working hour then the time taken to complete the bridge will be definitely more.
From all the above cases it can be concluded that each task/work can be represented with 4 variables
M = number of men
D = number of days
H = number of hours per day
W = amount of work
And the basic relationship among the variables is
MDH/W = Constant
In the above specified situation
M = 10 , D = 10, H =8 and W = 1
Now try to answer the following questions
1. How long will it take 8 men to complete the bridge working 6 hours per day?
2. If the bridge is required to be completed in 5 days then how many workers need to work 6 hour per day to complete the job
3. How many days it ll take for 10 men working 10 hours per day to build 3 similar bridges?
Answer 1: M1=8 , H1=6 and W1=1
Using the formula
MDH/W= M1D1H1/ W1
10 x 10 x 8 = 8 x D1 x 6
=> D1 = 100/6 = 50/3 days
Similarly try to solve the other two questions and discuss it in forum if you have any difficulty in doing so.
With the above mentioned concept you can solve any kind of MAN HOUR WORK problem but for few specific problems it is easier to attempt the problem in a different way .
If five men finish a work in 6 hours then 3 men will finish the same work in __________ many hours?
Method 1: This problem assumes that all men do the same amount of Work in an hour, and that they would do, say, twice as much work in two hours.
The first phrase: "Five men finish the work in 6 hours" is simply a
Statement of how much work is to be done. We know how many men; we know how many hours; but we are not told how much work. That is a clue that the problem demands us to figure out what sort of number, with what sort of units, can be attached to the word work.
Each men works for 6 hours and there are 5 men. So the total effort given is 30 man hour. Now that we know how much work three men must do, it is not so hard to figure out how long it will take them to do it
Simply divide 30 by 3 and the answer is 10 hour.
You can use the MDH/W formula to calculate it and the result ‘ll be the same. Here W and D are 1.
The first step is to find out what fraction each men contributes to the completion of the job.
5 men complete the job in 6 hour.
=> One man can complete the job in 30 hours.
=> One man does 1/30th of the job each hour.
=> 3 men ‘ll do 3/30(=1/10)th job in each hour.
=> The job can be done in 10 hours
(Read the above lines carefully and try to understand it thoroughly as it‘ll help you in solving many MHW problem)
Let’s try to solve one more problem with method 2
Question: Rama is able to do a job in 10 days working alone and Anil is able to do the same job in 20 days, working alone
(i). How long will they take to do the job if both of them work together ?
(ii)If they work alternately (provided Rama starts)?
Sol.(i) Rama completes the job in 10 days
=>Rama does 1/10th of the job in one day
Anil completes the job in 20 days
=>Anil does 1/20th of the job in one day
So total job done in one day (when both work together) = 1/10 + 1/20 = 3/20
Number of days taken by them would be 20/3
Sol (ii) When they work on alternate days, starting with Rama, it means that on first day, Rama works. On second day, Anil works, on third day, Rama works, and so on.
Therefore, in first two days, the fractional contribution to the amount of work is 3/20.
So in 6 two days they would complete 18/20th of the work (i.e 12 working days)
So total job left after 12th day = 1 – 18/20 = 1/10
And on 13 th day Rama would be working who completes 1/10th of the job in a day.
So the task would be completed on 12 + 1 = 13 th day.
Example 1: Let there be three men for a construction job. A completes the job in 30 days, B in 20 days and C demolishes the construction in 60 days, each person working alone. When will the construction be complete if
(a) they all work together
(b) they work on alternate day starting with A, then B and then C
Sol.(a) When they all work together, fractional work done on each day = 1/30 + 1/20 – 1/60 = 4/60 = 1/15. Therefore, they will take 15 days to complete the job.
(b)When they work on alternate days, then we observe that in the first 3 days, the fractional work done is 4/60 (from (a)). The greatest multiple of this fraction such that the fraction is less than 1 is 14 times this fraction, i.e., 56/60. Therefore, in 14 x 3 = 42 days, the fractional work done is 56/60 and the work left is 4/40 or 1/15. On 43rd day, A will do 1/30th of the work and we will be left with 1/30th work, which will be done by B in 20/30 days or 2/3rd of the day. Therefore, total number of days taken would be 43 2/3 days.
A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are fill pipes?
Let the number of fill pipes be ‘n'. Therefore, there will be 8-n, waste pipes.
Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fill pipes will fill 1/8th of the tank in an hour.
Hence, n fill pipes will fill n/8th of the tank in an hour.
Similarly, each of the waste pipes will drain the full tank in 6 hours. That is, each of the waste pipes will drain 1/6th of the tank in an hour.
Therefore, (8-n) waste pipes will drain ((8-n)/6)th of the tank in an hour.
Between the fill pipes and the waste pipes, they drain the tank in 6 hours. That is, when all 8 of them are opened, 1/6th of the tank gets drained in an hour.
(Amount of water filled by fill pipes in 1 hour - Amount of water drained by waste pipes 1 hour) = 1/6th capacity of the tank drained in 1 hour.
A takes 2 days to finish a task, B takes twice that amount of time, C takes twice of B, and D takes twice of C. If they do it in pairs, one pair takes twice the amount of time as the other,, which is the pair that takes longer?
a. A, B
b. B, C
c. B, D
d. C, D
30 cows graze a piece of grassland in 60 days, whereas 40 cows graze in 40 days. In how many days will 20 cows graze the grass (assuming the grass grows every day)?
Sol.Let a be the initial quantity (in relevant units) of grass and b be the amount of grass eaten per day.
Then, 60 x 30 = 1800 = a + 60b and
40 x 40 = 1600 = a + 40b. Solving these equations, we get a = 1200 units and b = 10 units.
If A and B work together, they will complete a job in 7.5 days. However, if A works alone and completes half the job and then B takes over and completes the remaining half alone, they will be able to complete the job in 20 days. How long will B alone take to do the job if A is more efficient than B?
Let a be the number of days in which A can do the job alone. Therefore, working alone, A will complete 1/a of the job in a day.
Similarly, let b the number of days in which B can do the job alone. Hence, B will complete 1/b of the job in a day.
Working together, A and B will complete (1/a + 1/b) of the job in a day. The problem states that working together, A and B will complete the job in 7.5 or 15/2 days. i.e they will complete 2/15th of the job in a day.
(1/a + 1/b)=2/15 -----(i)
From statement 2 of the question, we know that if A completes half the job working alone and B takes over and completes the next half, they will take 20 days.
As A can complete the job working alone in ‘a’ days, he will complete half the job, working alone, in a/2 days.
Similarly, B will complete the remaining half of the job in b/2 days.
Therefore, or a/2 + b/2 = 20
=> a + b = 40
=>a = b- 40 ------- (2)
From (1) and (2) we get,
=> 600 = 80b - 2b2
=> b2 - 40b + 300 = 0
=> (b - 30)(b - 10) = 0
=> b = 30 or b = 10.
If b = 30, then a = 40 - 30 = 10 or
If b = 10, then a = 40 - 10 = 30.
As A is more efficient then B, he will take lesser time to do the job alone. Hence A will take only 10 days and B will take 30 days.
If ‘A’ takes 10 days to do a job, he will do 1/10th of the job in a day. Similarly, if 2/5ths of the job is done in a day, the entire job will be done in 5/2 days
Mark can dig a ditch in 4 hours. Greg can dig the same ditch in 3 hours. How long would it take them to dig it together?
Solution: Let x = number of hours to dig the ditch together. If Mark takes 4 hours to dig the ditch, he can dig 1/4 of it in 1 hour. Greg can dig 1/3 of it in one hour. Mark's rate is 1/4 and Greg's rate is 1/3. If it takes them x hours to dig it together, they can dig 1/x part of it in 1 hour together. The total of the fractional part each can dig or 1/3 + 1/4 = the fractional part they can dig together in 1 hour.
1/3 + 1/4 = 1/x
Multiply be 12x to clear the fractions.
4x + 3x = 12
7x = 12
x = 12/7 or 1 5/7
1. To solve work problems, you need to work with the same unit of measure within each problem. For example, you cannot mix hours and minutes in the same equation.
2. You need to find the fractional part of the job that would be done in one unit of time, such as 1 minute or 1 hour. If a person can do a complete job in 3 days, he can do 1/3 of it in 1 day.
3. The fractional part of the job one person can do in 1 day plus the fractional part another person can do in 1 day equals the fractional part of the job the two can do together in 1 day. Example: If Bill can build 1/3 of a dog house in 1 day and Gary can build 1/5 of it in 1 day, together they can build 1/3 + 1/5 of the dog house in 1 day.
4. Rate of work x time = work done.
5. MDH/W = Constant
Now try to solve the problems in question bank