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Arithmetic
Time And Work  1
Time And Work
The chapter on time and work constitutes an important portion of the Arithmetic section. If one goes through the previous years’ papers on quantitative aptitude in any competitive exam, then surely he will come across questions on this chapter. It involves the basic tools of percentage, ratio and proportion ,fractions and a conceptual understanding of how to apply these tools. Once u get familiar with all the models and the type of questions asked, this could prove to be a very scoring chapter. So, let’s start with the basics.
If I say that Raghu can complete a work in 20 days, then there is a very strong underlying assumption that Raghu works at a constant rate i.e. he completes equal parts of the work every day. Now this work could be anything from painting a house or digging a well or building or road or anything you can think of. Now if we assume that the work is equivalent to 1part or 1 unit, then we can surely say that Raghu completes 1/20 parts or 1/20 units in one day.
Now let’s assume that Raghu has a friend by the name Rajiv and he can complete the same work in 30 days. This implies that he can complete 1/30 units or parts in 1 day. ( Again we have to assume that Rajiv also works at a constants rate).
If both of them were to work together then, in 1 day Raghu would have done 1/20 work and Rajiv would have completed 1/30 work.
Thus, in 1day they would have together completed 1/20 + 1/30 = 1/12 units.
Thus, if they work together for 12 days, then the entire work can be completed.
This could be approached in a slightly easier manner. Instead of assuming the total work to be 1 units, we can assume the total work to be the L.C.M. of 20 and 30 (20 and 30 are the number of days taken by Raghu and Rajiv to finish the work by working alone) i.e. 60.
Now as Raghu completes 60 units of work in 20 days, so he completes 3 units per day and similarly Rajiv completes 2 units of work each day. So, together in a day they complete 5 units of work and thus the number of days required to finish 60 units of work comes out to be 12.
The best part of this approach is that we don’t have to deal with fractions and so the calculations are quite easy to perform.
This approach can be extended even to three or four or as many persons we want and not just limited to two persons.
Example 1: A can complete 1/3 of a work in 5 days and B, 2/5 of the work in 10 days. In how many days both A and B together can complete the work?
Solution: As A can complete 1/3 rd of the work in 5 days, so he can complete the entire work in 5*3 = 15 days.
Similarly B can complete 2/5 of the work in 10 days, so he can complete the entire work in 10 * (5/2) = 25 days.
Now let the total work be 75 units ( L.C.M. of 15 and 25)
So, A completes 5 units in one day, B completes 3 units in one day and together they complete 8 units every day.
Thus, total work can be completed in 75/8 days i.e. 938 days.
Example 2: Ronald and Elan are working on an Assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

7 hrs 30 minutes 3. 8 hrs. 15 min.

8 hrs 4. 8 hrs. 25 min.
Solution: Ronald can type 32/6 i.e. 16/3 pages in 1 hour while Elan can type 40/5 i.e. 8 pages in 1 hour.
Together they can type 16/3 + 8 = 40/3 pages in 1 hour
So time taken by them to type 110 pages = 110/(40/3) = 33/4 hours or 8 hrs 15 min
Example 3: A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

8 2. 6 3. 4. 5
Solution: Let the total units of work be 90 units ( L.C.M. of 18 and 15)
Units completed by A in 1 day = 90/18 = 5 while units completed by B in 1 day = 90/15 = 6
Given that B worked on it for 10 days, so units completed by B = 10* 6 = 60
Units of work left = 90 60 =30
Now these 30 units can be completed by A in 30/5 = 6 days ( As A can finish 5 units per day)
Example 4: If A works alone, he would take 4 days more to complete the job than if both A and b worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?

10 days 2. 12 days 3. 6 days 4. 8 days
Solution: Here we don’t have a clue about how many days are taken by A and B working alone, so taking L.C.M. would not be an option. In fact this is a special case for which there is a standard result. Let us try to derive that.
Let both A and B can complete a work working together in T days
And let the number of days taken by A to finish the work alone is x more than the number of days taken when both were working together. Similarly let the number of days taken by B to finish the work alone is y more than the number of days taken when both were working together.
So, 1 day work of A is 1/(T+x) and similarly 1 day work of B is 1/(T+x)
Example 5: A can complete a work in 20 days and B in 30 days. A worked alone for 4 days and then B completed the remaining work along with C in 18 days. In how many days can C working alone complete the work?
Solution: Let the total work be 60 units (L.C.M. of 20 and 30).
Units completed by A in 1 day = 60/20 = 3 and units completed by B in 1 day = 60/30 = 2
As A worked alone for 4 days. So, units completed by him = 4*3 = 12.
Units left = 6012 = 48
Let the number of units completed by C in 1 day = x
So number of units completed by both B and C working together in 18 days = 18(2+x)
So, 18(2+x) = 48 or x = 2/3 units.
So, number of days required by C to finish 60 units = 60/(2/3) = 90
Example 6: A and B together can do a work in 10 days. B and C together can do the same work in 6 days. A and C together can do the work in 12 days. The, A,B and C together can do the work in
Solution: Let the total work be 60 units ( L.C.M. of 10,6 and 12)
Sum of units completed by A and B in 1 day = 60/10 = 6 i.e. A + B = 6
Sum of units completed by B and C in 1 day = 60/6 = 10 i.e. B + C = 10
Sum of units completed by A and C in 1 day = 60/12 = 5 i.e. A + C = 5
Adding them, we get 2(A +B+C) = 21
Or A + B +C = 21/2
So, in 1 day A ,B and C together can complete 21/2 units. So, time taken by them to complete 60 units = 60/(21/2) = 557 days
This problem can also be solved using fractions
1/A + 1/B = 1/10 (1) , 1/B + 1/C = 1/6 (2) and 1/A + 1/C = 1/12 (3)
2(1/A + 1/B + 1/C) = 21/60 or (1/A + 1/B + 1/C) = 21/120
So, time taken by A, B and C to finish the work = 120/21 = 557 days.
In the same questions if we were to find the number of days taken by C alone to finish the work, then
A+B+C = 21 i.e. A,B and C together complete 21 units in 1 day
Also A+B = 6 , so C = 15, i.e. C can complete 15 units alone in 1 day
So, time taken by C to complete 60 units = 60/15 = 4 days
Similarly we can also find out the number of days taken by A or B working alone.
Example 7: A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in 

15 days 2. 20 days 3. 25 days 4. 30 days
Solution: Let the total work be 50 units (L.C.M. of 10 and 50)
So A+B = 50/10 = 5 (i.e. A and B together can complete 5 units in 1 day) (i)
Similarly C = 50/50 = 1 (i.e. C can complete 1 unit in 1 day) (ii)
Also the number of units completed by A in 1 day =Units completed by both B and C in 1 day
So, A = B +C (iii)
From (i), (ii) and (iii), we get B = 2 units
So, time taken by B alone to finish the work = 50/2 = 25 days.
Working Alternately:
Sometimes questions are asked if the persons work alternately and not together and moreover they can surprise u with all sorts of twists and variations. There is no formula or a short trick that u need to remember. Simply apply the concept that we have already studied.
Example 8: A, B and C can complete painting a house in 12, 20 and 30 days respectively by working alone. In how many days can the work be finished if
a. They work on alternate days starting from A, then B, then C and then A again and so forth.
b. On the first day both A and B work together, on the second day both B and C work together, on the third day both A and C work together and this cycle is repeated till the work is completed.
c. A works on all the days, B works on every alternate day and C works on every third day. On the first day they all worked together.
d. A works on all the days while B and C join him on every alternate day starting from the first day.
Solution: Let the total work be 60 units ( L.C.M. of 12,20 and 30)
Units completed by A in 1 day = 5, units completed by B in 1 day 3 and units completed by C in 1 day = 2
a. Given that A works on the first day, then B on the second, then C on the third. So, in the first three days total units of work completed = 5+3+2 =10
Now we need to complete 60 units. It implies we need to have 6 such cycles of three days each. Thus, the work can be completed in 6*3 = 18 days.
b. Given that on the first day both A and B work and complete 5+3 = 8 units
On the second day both B and C work together and complete 3+2 = 5 units
On the third day both A and C work together and complete 5+2 = 7 units.
Thus on the first three days total units of work completed = 8 + 5 + 7 = 20
Now as 60 units are to be completed, we need three such cycles of three days each. So, time taken to complete these 60 units = 3*3 = 9 days.
Days 
Persons working 
Units completed 
1 
A, B and C 
5+3+2 = 10 
2 
Only A 
5 
3 
Both A and B 
5+3 = 8 
4 
Both A and C 
5+2 = 7 
5 
Both A and B 
5+3 = 8 
6 
Only A 
5 
7 
A,B and C 
5+3+2 = 10 
8 
Only A 
5 
Till the end of 8th day 58 units have been completed. Now the next day A and B will work together and can finish 5+3 = 8 units in that day. But we need only 2 units to be completed. So, these 2 units can be completed in ¼ day.
Concept of Efficiency :
Efficiency is related to the ability to perform a task with minimum efforts or minimum wastage of resources. It can be thought of as a ratio of output to the input provided in performing a task. It can be quantitatively determined. But we are not going to go into the deeper concepts and definitions involving efficiency and will study it in relation to this chapter.
Here we are going to define efficiency of one person in relation to the other. For e.g. if we say A is 50% more efficient than B, then it means that if B finishes 1 unit then in the same time A will complete 1.5 units. So if efficiency of B is e, then efficiency of A is 3/2 times e. Now definitely as A is more efficient so he will take lesser time to complete the same work when compared to B. In fact if B takes T time to finish a work, then A would take (2/3)* T to finish the same work ( As efficiency has inverse relation with time i.e. if the efficiency is more, then time taken is less and vice versa)
Example 9: X is 3 times as fast as Y and is able to complete the work in 40 days less than Y. Then the time in which they can complete the work together is
Solution: As X is 3 times as fast as Y so if Y takes T days to finish the work, then X will take T/3 days.
So, T – T/3 = 40 or 2T/3 = 40
Thus, T = 60 days
X can complete the work in 20 days while Y in 60 days. Let the total work be 60 units. So X can complete 3 units per day, while Y completes 1 unit per day.
Time taken to finish the work = 60/4 = 15 days { As they together complete 4 units in 1 day }
Example 10: A is 60% as efficient as B. C does half of the work done by A and B together. If C alone does the work in 20 days, then A, B and C together can do the work in
Solution: As A is 60% efficient as B, so if B completes a units in one day then A would complete 0.6a units in 1 day.
Let the total work be 20 units, so C completes 1 unit in 1 day.
As C does half the work done by A and B together, so we can say if C completes 1 unit in 1 day, then A and B together complete 2 units in 1 day
Thus, together they complete 3 units of work per day. So total time taken by them to complete 20 units = 20/3 = 623 days.
If we were to find out number of days taken by A and B alone to finish the work, then
Or 0.6a + a = 2 ( As they complete 2 units together in a day)
or a = 2/1.6 or a = 1.25 or 5/4 units
So, B completes 1.25 units in 1 day and A completes 1.25*0.6 = 0.75 or ¾ units
Number of days taken by B alone = 20/(5/4) = 16 days
Number of days taken by A alone = 20/(3/4) = 2623 days
Man Day Work /Man Hour Work
Example 11: A certain number of men can finish a piece of work in 100 days. However, if there were 10 men less, then it would take 10 days more for the work to be finished. How many men were there originally?

75 2. 50 3. 100 4. 110
Solution: Let initially there were x men involved and now there are x100 men.
Using the formula M1*D1 = M2*D2 { As the work remains constant }
x * 100 = (x10)*110
or x = 1.1x – 11 or x = 110
Example 12: A man undertakes to do a certain work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days. The number of additional men that should be appointed so that the whole work will be finished in time is:

50 2. 75 3. 100 4. 125
Solution: As the man employs 200 men and they work for 50 days to finish a quarter of work i.e. ¼th of work is finished. We need to find the additional men that would finish the remaining ¾ th of work in next 100 days ( As the total work needs to be finished in 150 days)
Example 13: 3 men and 4 women can complete a work a task in 4 days while 4 men and 2 women can complete the same task in 6 days. In how many days will 6 men and 2 women be able to finish the same work?
Solution
4(3men +4 women) = 6 (4men + 2 women)
12 men + 16women = 24 women + 12women
12 men = 4 women
3 men = women.
Let 1 man does 1 unit of work, so 1 woman does 3 units of work.
total work = 4(3*1+4*3)
= 4 (3+12)
= 60 units
no of units of work done by 6 men and 2 women in 1 day = 6*1+2*3
= 12 units
So, time taken = 60/12=5 days.
Example 14: 2 men, 3 women and 4 children together can complete a work in 4 days while 3 men, 3 women and 2 children can complete the same work in 3 days. In how many days will 5 women and 2 children finish the same work; if a woman is twice as efficient as a child?
Solution
4(2m + 3w + 4c) = 3(3m + 3w + 2c)
8m + 12w + 16c = 9m + 9w + 6c
m = 3w + 10c
also w = 2c.
m = 6c + 10c
m = 16c
Let 1 child completes 1 unit of work in 1 day, so 1 man completes 16 units of work in 1 day and 1 woman completes 2 units of work in 1 day.
Total work = 4(2*16+3*2+4*1)
= 4 (32+6+4)
= 168 units
5 women and 2 children can complete
5 * 2 + 2*1 = 12units of work in 1 day.
total time taken = 168/12 = 14 days.
Example 15: Thirty men can complete building a road in 6 days. On the first day all the men reported for work but on each subsequent day 2 men left the work. i.e on the second day 28 men worked, on the third day 26 men worked and so on . In how many days will the road be built?
Solution
Let each man completes 1 unit of work per day.
Total work = 30 * 6 = 180 units.
On the first day, 30 men reported , so units of work got completed on 1st day = 30.
on the second day, units of work completed = 28 and so on.
Let the men worked for n complete days