Work to be done usually considered as one unit. It may be constructing a wall or road.
Filling up or emptying a tank or cistern or eating certain amount of food.
There are some basic assumptions that are made in the problems on time and work.
These are taken for granted and are not specified in every problem.
1) If a person (or one member of the workforce) does some work in a certain number of days, then we assume (unless otherwise explicitly stated in the problem) that he does the work uniformly, ie he does the same amount of work everyday.
For example if a person can do some work in 15 days he does 1/15 day
If a person completes the work in 4 days, he does 1/4 th th of the work in one
of the work on each day and conversely, if a person can complete 1/4 th of the work in one day, he can complete the work in 4 days.
2) If there is more than one person (or members of workforce) carrying out the work, it is assumed that each person (or members of the workforce), unless otherwise specified, does the same amount of work each day. This means they share the work equally.
For example if a tap can fill a tank in 20 minutes than in cone minute, it can fill 1/20 part of the tank
If two people together can do the work in 8 days it means that one man can do it in 16 days. This is turn means, each person can do 1/16 th of the work per day.
If a man works three times as fast as a boy does, the man takes one third of the time the boy takes to complete the work. If the boy takes 12 days to complete the work then the man takes 4 days to complete the work.
This method is known as “UNITARY METHOD” ie the time taken per “UNIT WORK” or number of persons required to complete “UNIT WORK” or work completed by “unit person” in “unit time etc is what is first calculated.
We should recollect the fundamentals on variation (direct and inverse) here
The work and men are directly proportional to each other, ie if the work increases the number of men required increases to complete the work in the same number of days and vice versa.
Men and days are inversely proportional ie if the number of men increases, the number of days required to complete the same work decreases and vice versa.
Work and days are directly proportional ie if the work increases the number of days required also increases if the work is to be completed with same number of men and vice versa.
The concept of man days is very important and useful here. The number of men multiplied by the work number of days that they take to complete the work will give the number of man days required to do the work. The total number of Monday’s required to complete a specific task will remain a constant. So if other will change accordingly so that their product will remain constant.(remember from our knowledge of variation, two variables whose product is a constant are said to be inversely proportional to each other). The two variables men and days are inversely proportional to each other.
EXAMPLE PROBLEMS
1) If 20 men take 30 days to complete a job in how many days can 25 men complete the job?

Sol:
If 20 men can complete the job in 30 days then the number of mondays, required for the work is 20*30=600 mondays
If this work is now is to be done by 25 men since the no of mondays will be 600, the no of days they will take is 600/25=24 days
2) Fifteen men take 10 days to complete the job working 12 hours a day. How may hours a day should 10 men work to complete the job in 20days?
Sol:
Since 15 men take 10 days at 12 hours per day, the total men hours required for the job is 15*10*12=1800
When 10 men are required complete the same job in 20 days the no of men hours required for the job will still be the same. Hence the no of hours they should work per day is 15*10*12/10*20=9hours
Therefore 10 men can complete the work in 20 days working for 9 hours per day.
GENERAL FORMULA F1
Hence in general we say that
If m1 men can do w1 work in d1 days working h1 hours per day and m2 men can do w2 work in d2 days working h2 hours per day(where all men work at the same rate)then,
M1 D1 H1/W1= M2 D2 H2/W2
3)
A piece of work can be done by 16 men in 8 days working 12 hours a day. How many men are needed to complete another work, which is three times the first one, in 24 days working 8 hours a day
Sol:
In the formula above F1 M1 D1 H1/W1= M2 D2 H2/W2
We know that w1=1 and w2=3
M1=16, h1=12,d1=8,h2=8,d2=24
16*8*12/1=m1*8*24/3=m1 24 men
FORMULA F2
If two persons A and B can individually do some work in p and q days respectively, we can find out how much can be done by them together in one day. Since A can do 1/p work in one day and B can do 1/q work in one day, the two of them together do 1/p+1/q work in one day.
From this we can find out the no of days that they take to complete the work
If A can do a piece of work in p days and B can do it in q days then A and B together can complete the same in pq/p+q days.
4)
A can do a piece of work in p days, B can do the same work in 12 days, in how many days can the work be completed if the A and B work together?
Sol: A and B together can do the work in 9*12/9+12=108/21=5*1/7days
5)
A and B together can do 1/12
Sol: th of work in 12 days and A alone can complete the work in
18 days. How long will B alone take to complete the job?
In a day A and B together can do 1/12 th of the work in a day, A alone can do 1/18 th of the work
Therefore in one day work done by B alone is 1/12-1/18=1/36
So B alone can complete the work in 36 days.

6)
Anil and Amit can complete a work in 12 days working together. Amit alone can complete it in 16 days. Both of them worked together for 4 days and then amit left. How long will anil will take to complete the remaining work?
Sol:
Work done in 1 day by both together =1/12
Work done by both in 4 days =4*1/12=1/3
Remaining work =2/3
Work done by anil in 1 day =1/12-1/16=1/48
So anil alone takes 48 days to complete the work since only 2/3 of the work is left, anil will do the remaining work in 2/3*48=32days
7)
A and B together can do a piece of work in 12 days, B and C can do it in 15 days and C and A can do the same work in 20 days . How long would each take to finish the work?
Sol:
Work done by A and b in 1 day=1/12
Work done by B and C in 1 day=1/15
Work done by C and A in 1 day=1/20
Adding all three we get work done by 2(A+B+C) in 1 day =1/12+1/15+1/20=1/5
So A,B and C together finish in 1 day 1/10 th of the work
Work done by A in 1 day =work done by A,B and C in 1 day-work done by B and C in 1 day=1/10-1/15=1/30
So A alone can do it in 30 days
Work done by B in 1 day=1/10-1/20 =1/20
So B alone can do it in 20 days
Work done by C in 1 day=1/10-1/12=1/60
So C alone can do it in 60 days
8)
A and B can do work in 12 days B and C in 15 days and C and A in 20 days. They all work together for 6 days and then a left. In how may more days can B and C finish the remaining work?
Sol:
Work done by A, B and C in 1 day
=1/10(As calculated in previous sample)
Work done by A,B and C in 6 days= 6*1/10=3/5
So B and C can do the remaining 2/5 of the work in 2/5*15 = 6 days
9)
A can do a work in 12 days. When he had worked for 3 days B joined him. If they complete the work in 3 more days in how many days can be alone finish the work?
Sol:
Work done by A In 1 day =1/12
No of days A worked =3+3=6
So total work done by A=6*1/12=1/2
The remaining ½ of the work is done by B in 3 days
So complete work will be done by B in 3*2/1=6days
10)
A and B together can do a piece of work in 14 2/5days, B and C together can do the same work in 12 days. After A worked for 8 days, B for 12 days C takes up and finished it alone in 6 days. In how many days will each of them do the work, working alone?
Sol:
Work done by A and B in 1 day=5 7/12
Work done by B and C in 1 day =1/12
A worked for 8 days, B worked for 6 days to complete the work
8a+12b+6c=1 where

a,b,c are the workj done in one day by A,B,C respectively
(8a+8b)+(4a+4c)+2c=1
8(5/720+4(1/12)+2c=1
40/72+4/12+2c=1
2c=1-(64/72)=8/72 =4/72=1/18 so C alone can complete the work in 18 days work done by B alone in 1 day=(b+c)-c=5/72-1/36=1/24 so B alone can do it in 36 days and A alone can do it in 24 days.
11)
To do a certain work C alone takes twice as long as A and B together, A would take 3 times as B and C together. All the three complete the work in 5 days. How long would take separately?
Sol:
3 times A’s daily work= (B+c’s) 1 days work adding 1 time A’s daily work to both sides we get 4 times A’s daily work= (A+B+C’s)daily work=1/5
A’s daily work =1/20
2 times c’s daily work =(A+B)’s daily work adding 1 times C’s daily work to both sides we get 3 times C’s daily work
= (A+B+C)’s daily work =1/5
So C’s work =1/5; C takes 15 days to do the work
So B’s daily work= 1/5[1/20+1/15]=1/12
So a alone can do the work in 20 days, B alone in 12 days and C alone in 15 days
12)
4 men or 5 women can construct a wall in 82 days. How long will it take 5 men and 4 women to do the same?
Sol:
Given 4m=5w where m is work done by one man in one day and w is work done by one women in one day
=>1m=5w/4
Now 5m+4w=5[5w/4]+4w=41w/4if 5w can do the work in 82 days, 41w/4 can do in
5w*82*4/41w=40 days
13)
If 9 men and 12 boys can do a work in 4 days and 4men and 16 boys can do the same work in 6 days, how long will 6 men and 24 boys take to complete the same work?
Sol:
Given 9m+12b can do the work in 4 days and 4m+16b can do the same work in 6 days
So 4(9m+12b) =6(4m+16b) =>m=4b
Now 9m+12b=9(4m)+12b=48b
We want 6m+24b=48b
So 6 men and 24 boys also take 4 days to do the work.
14)
A certain no of men can do the work in 20 days. If there were 4 more men, the work can be done in 5 days less. How many men were initially?
Sol:
Let the initial no of men be p
No of days is 20; so no of man days is 20p
If there are 4 more men that is (p+4) it is completed in 15 days
So 20p=(p+4)15=>p=12
So initially there were 12 men
15)
X is three times as fast as Y and is able to complete the work in 40 days less then Y. find the time in which they can complete the work together?
Sol:

If Y does the work in 3 days, X does it in 1 day that is difference is 2 days.
But the actual difference is 40 days.
If the difference is 2 days, X takes 1 day and Y takes 3 days. If difference is 40 days
(that 20 times)
X takes 20 days Y takes 60 days
So the time taken together = 20*60/20+60 = 15 days
16)
Sita can do work in 12 days working 4 hours a day. Gita can finish the same in 15 days working 3 hours a day. How many days can they finish it working together at 4 ½ hours a day?
Sol:
No of hours taken by sita=12*4=48
No of hours taken by gita= 15*3 = 45
So together they will do
45+48/45*48=31/720 th work in one hour
Together they will take = 720/31 hours or 720/31 days working 1 hour a day so working 4 ½ hours a day they can complete the work in 720/31*9/2=720*2/31*9= 5
5/31 days
17)
A alone can do the work in 12 days and B alone in 18 days. If C takes twice as long as A and B together, how long wills B and C together take to complete the same work?
Sol:
A does the work in 12 days ; B does the same in 18 days. Hence, when they work together they take
12*18/12+18=36/5 days.
C take twice the time A and B together take
That is C take the 2*36/5=72/5 days to do the work
If B and C work together work done per day is = 1/18+5/72=9/72=1/8 hence B and C can do in 8 days
18)
A and B each working alone can do a work in 10 and 15 days respectively. They started the work together but B left after sometime and A finished the remaining work in 5 days.
After how may days form the start did B leave?
Sol:
A’s work for 5 days =5*1/0=1/2 work
The remaining half of the work was done by A and B together
Work done by A and B in a day is =1/10+1/15=1/6
So no of days they worked together is = ½ / 1/6 = 3 days
Hence, B left 3 days form the start of the work.
19)
A contractor decided to complete the work in 40 days and employed 60 men at the beginning 40 men additionally after 10 days and got the work completed as per schedule. If he had not employed the additional men how may extra days would he have to complete the work?
Sol:
100 men did the remaining work in (40-10)=30 days => 60 men did it in 50 days
So extra days needed is = 50-30 = 20 days
20)
A group of 35 men is employed to complete some work, in 48 days. After 33 days, 5 more men are employed and the work is finished 1 day earlier; if 5 more men were not employed how may more days would it have taken beyond the expected period?
Sol:
40 men will do it in 40*14/35=16 days

Extra time taken = 16-15=1 day behind the schedule
21)
A and B working separately can do the work in 6 and 12 day. They work on alternate days starting with A on the first day. In how many days will the work be completed?
Sol:
Since they are working on alternate days. Let us consider the time period of two days in which A does
One day’s and B does the next days work. in a period of two days work done by A and B together
=1/6+1/12=1/4
Since they complete 1/4 th of the work in a period of 2 days to complete the work they need 4 such periods of 2 days ie 4*2= 8 days
22)
A and B working separately and do a work in 6 and 9 days they work on alternate days starting with A on the first day. In how many days will the work be done?
Sol:
Since they are working on alternate days, let us consider a period of two days
If a period of two days, work done by A and B =1/6+1/9=5/18
If we consider 3 such time periods of 2 days (we are considering 3 periods because in the fraction of 5/18, the numerator 5 goes 3 times in the denominator 18)
Work done = 3*(5/18) =15/18
Remaining work = 3/18=1/6
Now it is A’s turn since 3 whole no of periods are over
Time taken by A to finish 1/16 th of the work is one day
So total time taken =3*2+1= 7 days
23)
A and B working separately can do a work in 12 and 15 days. They work on alternate days starting with A on the 1 st day. In how many days will he work be completed?
\
Sol:
In a period of 2 days work done = 1/12+1/15 = 3/20
In 6 such time periods of 2 days, ie 12 days (we are considering 6 periods because in the fraction 3/20, the numerator 3 goes to 6 times in the denominator 20)
Work done is 18/20
Remaining work = 1-18/20=2/20=1/10
Now it is A’s turn but a does only the 1/12
A works for one day (Ie the 13
Now it is B’ turn to do 1/60 th th of the work in 1 day. So balance work after th ) = 1/10-1/12=2/120=1/60
of work he takes 1/60 / 1/15 = 1/4 th of a day
Total no of days taken = (6*2) +1+1/4 = 13 ¼ days
24)
A and B working together separately can do a work in 20 and 24 days. They work on alternate days starting with B on the first day. In how many days will the work be completed?
Sol:
Work done by A and B in a period of 2 days= 1/20+1/24=11/120
If we consider 10 such tike periods (we are considering 10 periods because in the fraction 11/120 the numerator 11 goes 10 times in the denominator 120)
Work done = 10*11/120=11/12
Remaining work =1-11/12=1/12
Now it is B’s turn and he can complete 1/24 th of the work in 1 day so he cannot complete the work in 1 day. So he cannot complete the balance 1/12 th of the work. So after 1 days
Work done by B balance work=1/12-1/24=1/24
Now it is A’s turn and he can complete 1/20 th of the work in one day
So to complete 1/24 th of work he will take 20/24=5/6
Total tike taken = (10*2) +1+5/6 = 21 5/6 days th of the day.

NOTES
When the people doing some work earn some money together for doing the work then this money has to be shared by all the people doing the work.
In general money earned should be shared by people doing the work together in the ratio of the TOTAL WORK done by each of them.
For example if A 2/5 th of the work, then he should 2/5 work. In the remaining 3/5 th of the total earnings for the th of the work is done by B and C in the ratio 1:2 then the remaining 3/5 th of earnings(after paying A) should be shared by B and C in the ration of
1:2 suppose Rs 500 is paid to A,B and C together for doing the work, then A will get Rs
200(which 2/5 of 500)B will get Rs 100 and C Rs 200 (because remaining 300 after paying to A is to be divided in the ratio of 1:2 between B and C.
When people work for the same no of days each, then ratio of the total work done will be the same as the work done by each of them PER DAY. Hence if all the people involved work for the same no of days, then the earnings can directly be divided in the ratio of the WORK DONE PER DAY by each of them.
25)
A, B and c can do a piece of work in 4,5 and 7 days respectively. They got Rs 415 for the work. What is A’s share?
Since they work for the same no of days the ration in which they share the money is the ratio of the work done per day
That is ¼:1/5:1/7=35:28:20
Hence A’s share is (35/83)*415= Rs 175.
26)
A, B and c can do a work for Rs 4500. A and B together complete 3/5 th of the work and then C takes over the finished the work. What is the amount got by C?
Work done by C=1-3/5=2/5
So C’s amount =2/5*Rs 4500=Rs 1800
27)
Wages for 40 women and for 30 days are RS 21,600. How many men must work for 25 days to earn Rs14, 400 if the daily wages for a man is double that of a women?
Sol:
Wages of women per day = 21600/30*40= Rs 18
Wages of a man= 2*18 = Rs 36
So no of men = 14400/25*36 = 16
28)
A, B and C can together earn Rs 1,620 in 9 days. A and C can earn Rs 600 in 5 days; where as B and C in 7 days can earn Rs 910 find the daily earnings of C?
Sol:
Daily wages of A+B+B = 1620/9=180----- (1)
Daily wages of A+C = 600/5 = 120 ----- (2)
Daily wages of B+C = 910/7=130----- (3)
[(2) + (3)] ------ (1) gives
[A+B+2C]- [A+B+C] =C
= (120+130)-180=70
29)
Two men undertake a work for Rs 480; they can each do the work in 16 and 24 days. If they complete the work in 8 days with the help of a boy, how should they divide the money?
Sol:

The 1 st man can do 1/16 th work in one day and since he works for 8 days, he does
8*1/16=1/2th work. So he should get 1/2th amount =Rs 240
The 2 nd man can do the work in 1/24 th work in one day and he also worked for 8 days, he does 8*1/24=1/3th work. So he should get 1/3th amount = Rs 160
Since the 2 men earned Rs 400, the balance of Rs 80 will be paid to the boy.
PIPES AND CISTERNS
There can be pipes of the taps filling or emptying tanks with water. Times taken by different taps fill or empty the tank are different. Problems related to these can also be dealt with in the same manner as the foregoing problems on work have been dealt with.
These are only one difference between the problems on regular work. (of the type seen earlier)and those in pipes and cisterns. In pipes and cisterns, a filling pipe or tap does a positive work and an emptying pipe or a “leak” doe’s negative work.
30)
Two pipes A and B can fill a tank in 12 and 18 minutes. If both pipes are opened simultaneously, how long will it take to fill the tank?
Sol: part of tank filled by A in a minute= 1/12
Part of tank filled by B in a minute=1/18
Part of tank filled the pipes in one minute = 1/12
So the tank can be filled in 36/5=7 ½ minutes
+1/18= 5/36
31)
Pies A can fill a tank in 12 minutes and pipe B in 18 minutes and pipe C and empty a full tank in 36 minutes. If all of them work together, find time taken to fill the empty tank?
Sol:
Work done by the three pipes together in 1 minute = 1/12+1/18-1/36=1/9
So the empty tank will be filled in 9 minutes.
32)
Two pipes A and B can fill a tank in 20 and 30 minutes. There is an outlet C. if all 3 pipes are opened the tank will be filed in 24 minutes. How much time will it take for C alone to empty the full tank?
Sol:
Work done by c in one minute = (1/20+1/30)-1/24=1/12-1/24=1/24
So C can empty tank in 24 minutes.
33)
Two pipes A, B fill a tank in 20 and 30 minutes. if both pipes are opened at once, after how much time should A be closed so that the tank is filled in 15 minutes?
Sol:
Pipe woks for 15 minutes. In 1 minutes B fills 1/30 fills 15*1/30=1/2 th of the tank ie the 1/5 th minutes it
The remaining ½ is filled by A since A fills the tank fully in 20 minute. It takes 10 minutes to fill ½ of the tank. Hence A worked for 10 minutes. So A should be closed after 10 minutes.
34)
Three tapes A, B and C together can fill a tank in 3 hours. After 1 hour C is closed and the tank is filled in 4 more hours. Find the time in which C alone can fill the tank?
Sol:
Work of A+B+C IN 1 hour =1/3
Remaining part of the tank = 1-1/3=2/3
Time taken by (A+B)to fill this 2/3 rd of the tank = 4 hours
A and B together fill the tank in 6 hours
Now we know that A+B+C= 3 hours

A+B= 6 hours
So C=1/3-1/6=1/6
So C alone can fill the tank in 6 hours
35)
A tank has a leak, which would empty it in 8 hours. A tap is turned on which fills at the rate of 4 litres per minute and the tank is now emptied in 12 hours. Find the capacity of the tank?
Sol:
Work done by leak and filling tap in 1 hour =1/12
Work done by filling tap = 1/8+1/12+=1/24
Tank can be filled in 24 hours Capacity of tank = 24*60*4 =5760 litres