Time and Work Problems

Work and Time

This is an important chapter from the perspective of CAT and other entrance tests. It tests both logic and accuracy of calculations. It will use basic concepts of ratio and proportion in reference to speed and time. Students are advised to be highly attentive solving problems from this section.

Work is the job assigned or job completed, it will be same as distance here. The rate of work is the speed or speed of work.

Basics of Work and Time
1. Man × Days (or hours or minutes) = Man days, this is the basic formula to be understood:
Example 1: A man does a work in 10 days
(a)1 man × 10 days = 10 man days
(b)The work takes 10 man days
(c)Two men will take 5 days to finish the work (Total man days required are 10)
(d)The work done by one man in one day is 1/10th of total work

This is actually the only concept of work and time and one can solve the all the questions based on this concept. Even all the concepts are given below are derived from this concept of man days.

Example 2: Ram will do a piece of work in 15 days, what part of work will he do in two days?
Here Man × Days = Man days
1 × 15 = 15, as given work will take 15 days
Therefore in one man-day 1/15th of work will be done and in 2 days 2 × 1/15th = 2/15th of the work will be done.
2. B works twice as good as A, he will finish the work in half the days

Example 3: Ram is twice as good as Shyam in work; Shyam will do a piece of work in 30 days, in how many days ram will do the work?
Since Ram is twice as good, he will do the work in 30/2 = 15 days
3. If A and B can do a piece of work in X and Y days respectively while working alone, they will together take XY/(X + Y) days to complete it.
Proof:
A’s 1 day of work = 1/X
B’s 1 day of work = 1/Y
So (A + B)‘s total 1 day work
= (1/X + 1/Y) of the total work.
Let total work be W. Now (1/X + 1/Y) of W can be finished in 1 day
W (total work) can be finished in
=(1/1/x+1/y) = XY/X+Y days.

Example 4: Shyam will do a piece of work in 30 days; Ram can do same work in 15 days, in how many days can both do the work?
As per the formula, in days.
Therefore Therefore 30x 15 /30 + 15 = 450/50 = 10 days
4. If A , B , C can do a piece of work in X , Y , Z days respectively while working alone , they will together take XYZ /[ XY + YZ + ZX ] days to finish it

Example 5: Shyam will do a piece of work in 30 days; Ram can do same work in 15 days, Bhuvan can do the same work in 10 days, in how many days can all three do the work?
As per the formula, in days.
Therefore 30x 15 /30 + 15 = 450/50 = 5 days

PIPES AND CISTERNS
The Pipes and cisterns use the same principles as of time and work. Here a pipe connected with a cistern is called an inlet pipe to fill it or an outlet pipe to empty it.
Important points

1. If an inlet pipe can fill a cistern in A hours, the part filled in 1 hour = 1/A (same as work and time fundamental)
2. If pipe A is ‘x’ times bigger than pipe B , then pipe A will take 1/xth of the time taken by pipe B to fill the cistern.

Example 6: It take 4 hrs for pipe A to empty a 100 liter tank, if another pipe B which double the size of pipe is employed how long will it take to empty the tank?

Since the Pipe is double the size, it will take ½ time of the time taken by the smaller pipe
Therefore 1/2 × 4 = 2 hrs

3. If an inlet pipe can fill a tank in A hours and an outlet pipe empties the full tank in B hours, then the net part filled in 1 hour when both the pipes are opened
= 1/a - 1/b

In 1 hour, the part filled (or emptied) = 1/a - 1/b
Time required to fill or empty the tank = ab / a-b hours

Example 7: There are two pipes (inlet and outlet) attached with a tank of 1000 litres, the inlet pipe can fill the tank in 2 hrs, the outlet pipe can empty the tank in 4 hrs. What is the time required to fill the tank in case both are open? In one hour what part of tank will be filled?
Inlet pipe time to fill the tank = 2 hrs
Outlet pipe time to empty the tank = 4 hrs
Time to fill the tank = hours = 4x2/2 = 4 hrs
Net part filled/emptied in one hour = 1/a -1/b =1/2 - 1/4 = ¼ th of the tank, which is even obvious from the earlier result.
4. If X and Y fill/empty a cistern in ‘m’ and ‘n’ hours, then together they will take (mn/m+n) hours to fill/empty the cistern and in one hour (n+m/nm) th part of the cistern will be filled/ emptied.(same as time and work)

Example 8: There are two pipes attached with a tank of 1000 litres, Pipe A can fill the tank in 2 hrs, the Pipe B can fill the tank in 4 hrs. What is the time required to fill the tank in case both are open?
Inlet pipe time to fill the tank = 2 hrs
Outlet pipe time to empty the tank = 4 hrs
Time to fill the tank = (mn/m+n)hours = 4x2/2+4 = 1.33 hrs

5. If an inlet pipe fills a cistern in ‘a’ minutes and takes ‘x’ minutes longer to fill the cistern due to a leak in the cistern, then the time in which the leak will empty the cistern is given by a(1+a/x)

Example 9: There is a pipe attached with a tank of 1000 liters, the inlet pipe can fill the tank in 2 hrs, there is a leak in the tank due to which it takes 2 hrs more to fill the tank. In what time can leak empty the tank?
Time take by pipe to empty the tank = 2 hrs
Extra time taken due to the leak = 2 hrs
By the formula,
Time taken for leak to empty the tank = a(1+a/x)
Therefore = 4 hrs

SOLVED EXAMPLES
Q1. A man works for 5 hrs/day for 3 days to finish a work. A boy finishes the same in 5 days working 12 hrs/day. How many boys are required to do the job in the time taken by 1 man?
Ans. Number of hours taken by man = 5 × 3 = 15
Number of hours taken by the boy = 12 × 5 = 60
To do the job in 15 hrs (time taken by man), boys required = 60/15= 4 boys

Q2. A set of farmers can completely harvest a crop in 10 days. However 12 farmers fell ill. The remaining now can do the job in 15 days. Find the original number of farmers
Ans. Suppose number of farmers = X
X can do the job in 10 days = 10X
Since 12 fell ill, X – 12 could do the job in 15 days
= (X – 12)15
Since job is same
10X = (X – 12)15
X = 36 farmers

Q3. If 6 men or 10 women can do a job in 2 days, find how long 4 men and 9 women to do the same.
Ans. Here 6 men’s work = 10 women’s work
One man’s work = women’s work
Therefore when it is 4 Men + 9 women
It is 4 men + 9 × men = 9.4 men
Since 6 men can do the job in 2 day
1 man can do the job in 12 days
9.4 men can do the job in = 1.27 days

Q4. 20 men can clean 2 km of lawn in 120 days, how many men can clean 3 km of lawn in 60 days?
Ans. Since 20 men can do 2 km in 120 days, therefore for 2 km man days required are 120 × 20 = 2400
For 3 km, which is 1.5 times 2, man days required will be 2400 × 1.5 = 3600
If days are 60, men required will be 3600/60 = 60

Q5. 20 men can clean 2 km of lawn in 120 days, how many men can clean 3 km of lawn in 60 days?
Ans. Since 20 men can do 2 km in 120 days, therefore for 2 km man days required are 120 × 20 = 2400

For 3 km, which is 1.5 times 2, man days required will be 2400 × 1.5 = 3600
If days are 60, men required will be 3600/60 = 60

Q6. 10 goats and 10 cows eat grass of 5 acres in a certain time. How many acres will feed 20 goats and 10 cows for the same time, supposing a cow eats as much as 2 goats?
Ans. Since cows eat twice than goats, then 1 cow’s work = 2 goats work
So for first job, 10 goats and 10 cows = 15 cows
And fore second job, 20 goats and 10 cows = 20 cows
If 15 cows can eat 5 acres in certain time, then in same time 20 cows will eat 20/15 × 5 = 6.67 acres

Q7. An engineer undertook to do a certain work in 40 days and employed 60 men for it. In 15 days only 1/4th of the work was done. How many extra men should the engineer employ in order to complete the work in time?
Ans. Work to be done in 40 days, as per schedule, 1/4th of the work done in 15 days therefore the total work will be done in 60 days. Now man days required for the entire job are 60 × 60 = 3600. Man days required for the left 3/4th of job = 3600 × ¾ = 2700
Man days left with engineer = 25 × 60 = 1500
Excess man days required = 2700 – 1550 = 1200
Here days cannot be extended therefore men required = 1200/25 = 48 men

Q8. A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits after 10 days, in how many days will the project be completed?
Ans. A can do the job in 20 days, Contribution in one day = 1/20, in ten days = 1/2
B can do the job in 30 days, Contribution one day = 1/30, in ten days = 1/3
Level of job done in 10 days with both
= 1/2 + 1/3 = 5/6, job left = 1/6
Only B has to do the job, B will do 1/6th job in 5 days (1/30th in one day)
Total days = 10 + 5 = 15 days

Q9. A and B can do a piece of work in 21 and 24 days respectively. They started the work together and after some days A leaves the work and B completes the remaining work in 9 days. After how many days did A leave?
Ans. Suppose A worked for X days
A can do the job in 21 days, in one day 1/21th of the job, in X days = X/21th of job
B can do the job in 24 days, in one day 1/24th of the job, in X + 9 days = X + 9/24th of job
Now since the job is done + = 1
Solving X = 7 days

Q10. 5 men and 6 boys finish a piece of work in 4 days; 4 men and 3 boys in 6 days. In how many days would 3 men and 6 boys finish the same work?
Ans. Man days for job done first way
= 4 (5M + 6B) = 20M + 24B
Man days for job done second way
= 6(4M + 3B) = 24M + 18B
As the man days are same,
20M + 24B = 24M + 18B
2M = 3B
Therefore for 3men and 6 boys
= 3 men and 4 men = 7 men
From first way, 9 men (5 men + 6 boys) can do the job in 4 days
Then 7 men will do the job in 9 × 4/7 = 36/7 days

Q11. Three pipes A, B and C can fill an empty cistern in 20, 10 and 15 hours respectively working alone. A is opened, after 5 hours B is opened and after another 2 hours pipe C is also opened. For how many hours C will be opened now to fill the cistern completely?
Ans. 1 hour work of A = 1/20
1 hour work of B = 1/10
1 hour work of C = 1/15
A worked for 5 hours then (A + B) worked for 2 hours and then suppose that (A + B + C) worked for another ‘x’ hours.
5(1/20) + 2(1/20 + 1/10) + x(1/20 + 1/10 + 1/15) = 1
Solving:
X = 27/13 hours