☛ If a person completes a work in X days, then his /her one day work will be 1/X
Meaning, if his total work has X parts, he would have completed 1 out of those X parts, where X is the number of days.
Eg. If a person can complete a work in 12 days, he will complete 1/12th work in 1 day.
Conversely, if a day’s work of a person is 1/X , he will complete the work in X days.
Two People Working on a Job Together
Let us say, Ram completes a job in A days and Shyam takes B days to complete the same job.
The total time taken by them is calculated as follows:
Ram’s one day work = 1/A
Shyam’s one day work = 1/B
Total one day work = 1/A + 1/B = (A+B)/AB
Thus the work will be completed in AB/(A+B) days.
Eg. Bheem completes building a bridge in 4 days, and Hanuman completes building a bridge in 2 days. How many bridges will the complete together in 8 days?
Bheem’s one day work = 1/4
Hanuman’s one day work = 1/2
Their one day work together = 1/4 + 1/2 = 3/4
Therefore, they take 4/3 days to construct one bridge together.
Considering that they have 8 days,
They will build 8/(4/3) = 6 bridges.
Concept of Man-Days
The product of men working on a job and the days taken to finish the job is known as ‘man-days’.
This is the same concept as Inverse Proportion in Unitary Method. It is used to find the number of days 1 person will take to finish the job, given A people finish the job in X days.
If A people can complete a work in X no. of days, then one person will complete the work in AX of days.
Eg. Food supplies in fort guarded by 10 people will last for 10 days. If the no. of soldiers are increased to 25, how long will the supplies last?
The food supplies are available for 10 soldiers for 10 days.
Let 1 soldier eat 1 unit of food every day, then there must be 10 x 10 =100 units of food in the fort.
If 1 soldier were to finish it, the food would last for 100 days.
Given that the no. of soldiers is increased to 25. Every day 25 units of food will be consumed in the fort.
Thus, the food will last for only 100/25 = 4 days.
Eg. If 10 persons can complete 2/3 rd task in 20 days, how long will the work be completed if 2 men were to complete the remaining task?
Soln:. Given that 2/3rd task is completed in 20 days, the task will be completed in
20x 3/2 = 30 days.
10 men together can complete the task in 30 days.
If 1 man completes 1 unit of work a day, 300 units of work together form the task.
Already 2/3 x 300 units = 200 units of work are completed.
Remaining work = 100 units
2 persons will complete 2 units of work per day.
Thus, time taken to complete remaining work = 100/2
= 50 days
PIPES & CISTERNS
The pipes and cisterns problems are very similar to the time and work problems
☛If a tap can fill a tank alone in X hours, in 1 hour it will fill 1/X of the tank.
Eg. If a tank is filled by a tap in 2 hours, then in first hour, 1/2 of the tank will be filled.
The other half of the tank will be filled in the second hour.
☛ If two taps are opened at the same time and one tap fills the tank in X hours while the other tap fills the tank in Y hours. Time taken to fill the tank will be
= XY/(X+Y) hours.
Here both the taps are doing the work of filling up the tank.
Eg. If one tap fills the tank in 8 hours while the other tank fills the tap in 10 hours. Find the total time to fill the tank completely.
Tank filled in 1 hr = 1/8 + 1/10 = 18/80
Thus, the tank will be completely filled in 80/18 hours.
☛ If one cistern fills the tank in X hours while the other empties the tank in Y hours, the tank will be filled or emptied depending on the values of X and Y.
Eg. A tank is filled by a pipe in 3 hours while another tap, empties it in 5 hours. Will the tank be emptied or filled?
Here one tap is doing the work of filling the tank while the other tap is doing the work of emptying the tank.
Work done by filling tap in 1 hour =1/3
Work done by emptying tap in 1 hour = 1/5
Work done by filling tap > Work done by emptying tank in 1 hour. So, the tank will be filled
Had the time of emptying tap < Time to fill the tank, tank will be emptied.
In the example, total work done by both taps = 1/3 – 1/5 (Since one tap is filling and the other is working against it, ie. emptying)
Work done in 1 hour= 1/3 – 1/5 = 2/15
Therefore, work will be completed (Tank will be filled) in 15/2 hrs = 7.5 hours.