POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-TIME & WORK

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Apr 18, 20140 comments


Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.


Time and Work
1.     Work from Days:
If A can do a piece of work in n days, then A's 1 day's work =
  1
.
n

2.    Days from Work:
If A's 1 day's work =
1
,
then A can finish the work in n days.
n

3.     Ratio:
If A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1.
Ratio of times taken by A and B to finish a work = 1 : 3.

4.     No. of days = total work / work done in 1 day


5.     Relationship between Men and Work

More men         ------- can do ------->             More work

Less men         ------- can do ------->             Less work


6.     Relationship between Work and Time

More work        -------- takes------>               More Time

Less work         -------- takes------>               Less Time


7.     Relationship between Men and Time

More men        ------- can do in ------->          Less Time

Less men          ------- can do in ------->          More Time


8.     If Mpersons can do W1 work in D1 days and M2 persons can do W2 work in D2 days, then 
        

9.     If Mpersons can do W1 work in D1 days for h1 hours and M2 persons can do W2 work in D2 days for h2 hours, then


 Note:  If works are same, then M1D1h1 = M2D2h2 


10.  If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is



11.  If A can do a work in ‘x’ days and A + B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is



I think the single most useful formula for the topic Time and Work is
N1H1D1E1W2 = N2H2D2E2W1
Where:
N1 and N2 = number of person
H1 and H2 = Hours worked by per person per day (assumed constant)
D1 and D2 = days
E1 and E2 = Efficiency
W1 and W2= Amount of work done

1. 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. The efficiency of the second group is half that of the first group?
Solution –
N1H1D1E1W2 = N2H2D2E2W1
16*12*8*1*3 = N2*8*24*0.5*1
http://www.indiabix.com/_files/images/aptitude/1-sym-cparen-h1.gifN2 = (16*12*8*1*3)/ (8*24*0.5*1) = 48
So number of men required is 48.
Note – you can remove anything from formula is not given in the question. For example if the question would have been –
2.“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.”
The applicable formula would have been –
N1H1D1W2 = N2H2D2W1
Since nothing is mentioned about efficiency, we remove it from both sides.
1. Time taken to do a certain job by a certain number of workers
2. The change in the number of hours required to do a job if the number of workers is changed.
3. The number of hours required to do the job  by different workers if their speeds are different.
4. Problems on wages earned by various workers executing a certain work in proportion with the amount of work done by each worker
5. Work done by various people in alternate intervals
6. Time taken by pipes/taps/leakages to fill/empty tanks/cisterns.
There are certain points that are to be kept in mind while doing problems. They are:-
1. A person does the same amount of work every day(unless specified in the problem).
Note
a. If A can do a piece of work in ‘n’ days, then the amount of work done by A in one day is 1/n.
b. Conversely, if A’s 1 day’s work=1/n, then A can finish the work in n days.
2. If there are more than one person involved carrying out the work collectively, it is assumed that the working capacity of each person is the same unless specified in the problem.
In general we can say that if w1 work is done by m1 men working h1 hours per day in d1 days and w2 work is done by m2 men working h2 hours per day in  d2 days then
m1d1h1w2=m2d2h2w1
Generally, the following types of question are asked in examinations and here are few time work problems shortcuts
Types of Questions and usage of formulas
 METHOD 1

M1 persons do a work in D1 days and M2 persons do the same work in D2 days then
we have the equation
          M1*d1= M2* D2

METHOD 2

M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days then
we have (M1 D1)/W1 = (M2 D2)/W2

METHOD 3

if A can do a work in x days, B can do a work in y days then
A and B together (A+B) can do the same work in ( xy )/(x+y) days  
                                                                                      
METHOD 4

A alone can do a work in x days , A+B can do a work in y days then .
B alone can do the work in (xy)/(x-y)

METHOD 5
here A+B means A and B working together
if A working alone takes x days more than (A+B) and B working alone takes y days more than (A+B) then
the number of days taken by A and B together is given by
        root of (xy)

METHOD 6

here A+B+C means A B C working together
A can do a work in x days , B can do the same work in y days , C can also do the same work in z days then
A+B+C can do the work in (xyz)/(xy+yz+zx) days

METHOD 7

here A+B+C means A B C working together
A+B can do a work in x days , B+C can do the work in y days, C+A can do the work in z days then
A+B+C can do the work in (2xyz )/xy+yz+zx days  
                                             
METHOD 8

if x1 men or y1 women can reap the field in D days ,then
x2 men and y2 women take to reap it in D(x1y1)/(x2 y1+x1 y2) days              
                                                        
 METHOD 9

a certain men can do a work in D days if there are x men less it could be finished in d days more, then
 the no of men orginaly are x(D+d)/d          
      
METHOD 10

a certain men can do a work in D days if there are x men more it could be finished in d days more,then
 the no of men originally are x(D-d)/d  
                 
METHOD 11

if x1 men or x2 women or x3 boys can do a work in D days then
 1 men + 1 women + 1 boy can do the same work in  d(x1x2x3)/(x1x2+x2x3+x3x1)
                      

     1.       A can do a work in 10 days. What is A’s 1 day work?

Sol: 1/10th of work

     2.       Two people can do a work in 10 days. What is 1 day 1 man work?

Sol: Since it is given as 2 people can do a work in 10 days, total work is 10 * 2 = 20 units of work.
Therefore, one day one man work = 1/total work = 1/20 units of work

      3.       Three men can do a work in 15 days. One man can do a work in how many days?

Sol: Since it is given as 3 people can do a work in 15 days, total work is 15 * 3 = 45 units of work.
Therefore, one man can complete the work in 45 days

      4.       One man one day work is 1/25. In how many days, can he finish the work?

Sol:  Number of days required = 1/(1/25) = 25 days
      
     5. A can work on 1km railway track in 1 day. In how many days, will he able to complete the work on 12km railway track?

Sol:     no. of days = total work / work done in 1 day
Therefore, no. of days taken = 12/1 = 12 days

      6. A can complete the work in 15 days. What fraction of work will be completed in 1 day?

Sol:      Let the total work is 1 unit.
Work in 1day = total work/no. of days to complete
                                    = 1/15th of work

      7.A can do a piece of work in 3 days and B can do a piece of work in 5 days. In how many days will the work be completed if both A and B work together?

Sol:
            Work done by A in 1 day = 1/3
            Work done by B in 1 day = 1/5
            Total work done by A and B in 1 day = 1/3 + 1/5 = 8/15
            Therefore, no. of days to complete work by A and B together = 1/(Total work) = 1/(8/15) = 15/8 days which is less than 3 and 5

             Using shortcut

            Let us consider the total work be 15 units (LCM of 3 and 5)
            So work done by A in 1 day = 15/3 = 5 units
            Similarly work done by B in 1 day = 15/5 = 3 units
            So total work done by A and B in 1 day = 5 + 3 = 8 units
Therefore, no. of days to complete total work i.e. 15 units = total work/work done in 1 day = 15/8 days
       
      Note:
                a.     Work done by A and B in 1 day will always be greater than that of A and B individually
          b. No. of days taken by A and B together will always be less than that of A and B individually

     8.A can do a piece of work in 6 days, B can do a piece of work in 4 days and C can do a piece of work in 12 days. Find the no. of days to complete the work if A, B and C work together?

Sol:
            Work done by A in 1 day = 1/6
            Work done by B in 1 day = ¼
            Work done by C in 1 day = 1/12
            Total work done by A, B and C in 1 day = 1/6 + ¼ + 1/12 = 12/24 = 1/2
            Therefore, no. of days to complete work by A, B and C together = 1/(Total work) = 1/(1/2) = 2 days which is less than 4, 6, 12

            Using shortcut
            Let us consider the total work be 24 units (LCM of 4, 6, 12)
            So work done by A in 1 day = 24/4 = 6 units
            work done by B in 1 day = 24/6 = 4 units
            work done by C in 1 day = 24/12 = 2 units
            So total work done by A, B and C in 1 day = 6 + 4 + 2 = 12 units
Therefore, no. of days to complete total work i.e. 24 units = total work/work done in 1 day = 24/12 = 2 days


      9.A can do a piece of work in 6 days and B can do a piece of work in 12. Find the no. of days to complete the work if A and B work alternatively?

Sol:
            Work done by A in 1 day = 1/6
            Work done by B in 1 day = 1/12
            Total work done by A and B working 1 day each = 1/6 + 1/12 = 3/12 = ¼
            Therefore, 1/4th of work is done in 2days.
No. of days to complete total work if A and B work alternatively = 1/((1/4)/2) = 8 days

            Using shortcut
            Let us consider the total work as 12 units (LCM of 6, 12)
            So work done by A in 1 day = 12/6 = 2 units
            work done by B in 1 day = 12/12 = 1 unit
            Total work done by A and B working 1 day each = 2 + 1 = 3 units in 2 days
            Therefore, work done in 1 day = work/no. of days = 3/2 units
            No. of days to complete work = total work/work in 1 day = 12/(3/2) = 8 days

      10.30 men can complete a job in 40 days. Then 25 men can complete the same job in how many days?

Sol: As per the formula M1D1 = M2D2
              30 * 40 = 25 * X = 30 * 40/25 = 48 days
     11.30 men can complete 1500 units in 24 days working 6hrs a day. In how many days can 18 men can complete 1800 units working 8 hrs a day?

Sol: As per the formula M1D1h1/W1 = M2D2h2/W2
          => 30*24*6/1500 = 18*x*8/1800
          => x = 36 days

      12.  A and B can do a work in 10 and 15 days respectively. Then combined A & B, in how many days the work will be completed?

      Sol: As per the formula x*y/(x + y)
               
A and B together can complete the work in 10 * 15/(10 + 15) = 6 days
     13.  A can do a work in 10 and, A and B together can do a work in 6 days. In how many days B can complete the same work?

      Sol: As per the formula x*y/(x - y)               B alone can complete the work in 10 * 6/(10 - 6) = 15 days

      14.  A is twice faster than B and B can complete in 12 days alone. Find the number of days to complete if A and B together work?

            Sol: Given B works in 12 days
            A is twice faster than B => A takes 2 times less time than B
            Therefore, A completes work in 12/2 = 6 days
            A and B together can complete in 12 * 6/(12 + 6) = 4 days

15. If 10 men or 18 boys can do a piece of work in 15 days, then 25 men and 15 boys together will do twice the work in:

Sol:10 men = 18boy hence 1 man = 18/10 boys
25 men + 15 boys = (25 * 18/10) + 15 = 60
now more work more days
more boys less days1 * 60 * x = 2*18*15 or x = (2*18*15)/60 = 9 days

16.A is thrice as good a workman as B and takes 10 days less to do a piece of work than B takes. B alone can do the whole work in

Sol: Ratio of times taken by A and B = 1:3.
If difference of time is 2 days, B takes 3 days.
If difference of time is 10 days, B takes (3/2) * 10 =15 days.

Sol: Let 1 mans 1 days work = x and 1 womens a days work = y
then 10x = 1/20 and 20y = 1/15 so x = 1/200 and y = 1/300
(5 men + 10 women)s 1 days work = (5/200) + 10/300 = 7/120.
So, they will finish the work in 120/7 = 17 and 1/7 days.

17.A is twice as good a workman as B and ‘together they complete a work in 15 days. In how many days can the work be completed by B alone

(As 1 days work):(Bs 1 days work) = 2: 1.
(A+B)s 1 day’s work= 1/15.
Divide 1/15 in the ratio 2 : 1
Bs 1 days work = (1/15) * (1/3) = 1/45
B alone can finish the work in 45 days. 

18. If Ramesh, Suresh and Harish can do a piece of work in 15 days, 10 days and 6 days respectiVelY how long will they take to do it, if all the three work at it together ? 

(Ramesh + Suresh + Harish)S 1 days work = [(1/15) + (1/10) + (1/6)] = 10/3 =1/3
So, all the three will finish the work in 3 days.

19.A and B can do a piece of work in 72 days, B and C can do it in 120 days. A and C can do it in 90 days. In what time can A alone do it?.

Sol: (A+B)s 1 days work = 1/72,
(B+C)s 1 days work = 1/120,
(A+C)s 1 days work= 1/90.
Adding 2(A+B+C)s 1 days work [(1/72) + (1/120) + (1/90)] = 12/360 = 1/30
(A+B+c)s 1 day’s work = 1/60.
As 1 days work = (1/60) - (1/120) = 1/120
A alone can fini the work in 120 days.

20. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do 

Sol: Suppose B takes x dáys to do the work.
A takes (2*(3/4)*x) = 3x/2 days to do it.
(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30.

21.45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?.

Sol: (45 * 16) men can complete the work in 1 day.
1 mans 1 days work = 1/720
45 mens 6 days work; = (1/16) * 6 = 3 / 8
Remaining work = (1 - (3/8)) = 5/8
75 mens 1 days work = 75/720 = 5/48
Now 5/48 work is done by them in 1 day.
5/8 work is done by them in (48/5) * (5x8) = 6 day

22. A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work was 

  Sol: (A+B)s 1 days work = (1/45) + (1/40) = 17/360.
Work done by B in 23 days =(1/40) * 23 =.23/40
remaining work = (1 — 23/40) = 17/40 .
Now, work was done by (A + B) in 1 day
17 /40 work was done by (A + B) in (1 * (360/17) * (17/40))= 9 days.
A left after 9 dáys.

23.4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it?

Sol: Let 1 mans 1 days work = x and 1 womans 1 days work = y.
Then, 4x + 6y = 1/8 and 3x+7y = 1/10
solving, we get y = 1/400
10 womens 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days.

24.12 children take 16 days to complete a work which can be completed by 8 adults in 12 days. 16 adults started working and after 3 days 10 adults left and 4 children joined them. How many days will it take them to complete the remaining work?

Sol: Let 1 child’s 1 day’s work = x and 1 adults 1 days work = y.
Then, 12x = 1/16 or x = 1/192 and 8y= 1/12 or Y = 1/196. Work done in 3 days = (16 * (1/96)*3)= 1/2
Work left = (1 - (1/2)) = 1/2
(6 adults + 4 children)s 1 days work = (6/96) + (4/192) = 1/12
1/12 work is done by them in 1 day.
1/2 work is done by them in (12 * (1/2)) = 6 days. 

25. Twelve men can complete a work in 8 days. Three days after they started the work, 3 more men joined. In how many days will all of them together complete the remaining work? 

Sol: 1 mans 1 days = 1/96
12 men’s 3 days work = (1/8) * 3 = 3/8.
Remaining work 1 — (3/8) = 5/8 15 mens 1 days work = 15/96
Now, 15/96 work is done by them in 1 day
5/8 work is done by them in (96/15) * (5/8) = 4 days

26. A can do of the work in 5 days and B can do of the work in 10 days. In how many days both A and B together can do the work?

 Sol: Whole work will be done by A in (5 x 3) = 15 days.
Whole work will be done by B m (10 * (5/2)) = 25 days.
As 1 day’s work = 1/15 and Bs 1 days work = 1/25
(A+B)s 1 daYs work=((1/15) + (1/25) ) = 8/75
So, both together can finish it in 75/8 = 9 and 3/8 days days

27.A and B can do a Piece of work in 5 days ;B and c can do it in 7 day; A and C can do it in 4 days. Who among these will take the least time if put to do it alone? 

Sol:(A+B)s 1 days work = 1/4
and (A+C)’s 1 day’s work = 1/4
2(A+B+C)s 1 days work = (1/5 + 1/7 + 1/4) = 83/140
(A+B+C)s 1 days work = 83/280
Cs 1 days work = (83/280 - 1/5) = 27/280
As 1 days work = (83/280 - 1/7) = 43/280
Bs 1 days work = (83/280 - 1/4) = 13/280
Thus time taken by A, B, C is 280/43 days, 280/13 days, 280/27 days respectively.
Clearly, the time taken by A is least.

28. A does half as much work as B and C does half as much work as A and B together. If C alone can finish the work in 40 days, then together all will finish the work in: 

 Sol: C alone can finish the work in 40 days.
(A + B)can do it in 20 days
(A + B)s 1 days wok = 1/20.
As 1 days work : Bs 1 days Work = 1/2 : 1 = 1:2.
A’s 1 day’s work = (1/20) * (1/3) = (1/60). [Divide 1/20 in the raio 1:2] Bs 1 days work = (1/20) * (2/3) = 1/30
(A+B+c)S 1 day’s work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40
All the three together will finish it in 40/3 = 13 and 1/3 days.

29. A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by B and C on alternate days? 

 Sol: (A+B)s 1 days Work = (1/11) + (1/20) = 31/220
(A+C)S 1 days work = (1/11) + (1/55) = 6/55
Work done in 2 days = (31/220) + (6/55) = 55/220 = 1/4
Now, 1/4 work is done in 2 days.
Whole work will be done in (4 x 2) = 8 days.

29. A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. The two together could complete the work in 

Sol: Work done by A in 10 days = (1/80) * 10 = 1/8
Remaining work=(1 — (1/8)) = 7/8
Now, 7/8 work is done by B in 42 days
Whole work will be done by B in 42 * (8/7) = 48 days.
As 1 days work = 1/80 and Bs 1 days work = 1/48
(A + B)’s 1 days work = [(1/80) + (1/48)] = 8/240 = 1/30
so, both will finish the work in 30 days.

30.A and B can separately do a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which B was replaced by C. If the work was finished in next 4 days, then the number of days in which C alone could do the work will be

Sol: (A + B)s 6 days work = 6 [(1/20) + (1/15)] = 7/20
(A+C)s 4days work = (3/10)*(A+C)s 1days work = 1/20 Cs 1 days work = (3/40) - (1/20) = 1/40
Hence, C alone can finish the work in 40 days.

31.A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together 8*(2/5) hours a day?
Sol:A can complete the work in (7 * 9) hrs = 63 hrs.
B can complete the work in (6 * 7) hrs = 42 hrs.
As 1 hours work = 1/63 and Bs 1 hours work = 1/42
(A+B)s 1 hours work = [(1/63) + (1/42)] = 5/126
Both will finish the work in (126/5) hrs
Number of days of8*(2/5) hrs each = (126/5) * (5/42) = 3 days.

32. A father can do a job as fast as his two sons working together. If one son does the job in 3 hours and the other in 6 hours, how many hours does it take the father to do the job’?

Sol: Fathers 1 hours work= (1/3) + (1/6) = 1/2
Time taken by father to complete the work = 2 hours.

33. A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in : .

Sol: Bs 5 days work = ((1/10) * 5) = 1/2
Remaining work = (1-(1/2)) = 1/2
A can do 1/2 work in 7and half days.

34. 12 men can complete a work in 18 days. Six days after they started working, 4 men joined them. How many days will all of them take to finish the remaining work?

Sol: (12 * 18) men can complete the work in 1 day.
1 mans 1 days work = 1/216
12 men’s 6 days work = (1/18)*6 = 1/3
remaining work = 1 - (1/3) = 2/3
16 mens 1 days work = 16/216 = 2/27
2/27 work is done by them in 1 day.
2/3 work is done by them in (27/2)*(2/3) = 9 days.

35. A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. A alone can finish the work in:.

Sol: Work done by A and B in 20 days = (1/30 * 20) = 2/3
Remaining work= ( 1 -2/3) = 1/3 Now,1/3 work is done by A in 20days
Whole work will be done by A in (20x3) = 60 days.

36. A can do a piece of work in 30 days while B alone can do it in 40 days. In how many days can A and B working together do it?

Sol:As l days work = 1/30 and Bs l days work= 1/40
(A + B)s 1 days work = (1/30 + 1/40) = 7/120
Both together will finish the work in 120/7 = 17.14 days.

37. A piece of work can be done by 6 men and 5 women in 6 days or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in 

Sol: Let 1 mans 1 days work =x and 1 womans 1 days work = y. Then, 6x+5y = 1/6 , 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90
(9 men + 15 women)s 1 days work = (9/54) + (15/90) = 1/3
They will finish the work in 3 days.

38. A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. The number of days taken in completing the work was

Sol: Work done by A in l0 days = (1/25) *10 = 2/5.
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100
9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (100/9) * (3/5) = 20/3days.
Total time taken = (10 + 20/3) = 16 * (2/3) days.

39.Ram can do a piece of work in 8 days which Shyam can finish in 12 days. If they work at it on alternate days with Ram beginning, in how many days, the work will be finished?.

Sol: (Ram + Shyam)s 2 days work = (1/8) + (1/12) = 5/24
Their 8 days work = (5/24) * 4 = 5/6
Their 8 days work = (5/6) + (1/8) = 23/24
Remaining work = (1 - (23/24))
Now it is Shyam’S turn.
1/12 work is done by him in 1 day.
1/24 work is done by him in (12 * (1/24)) = 1/2 day.
Total time taken = 9 and half days. 

40. A, B and C are employed to do a piece of work for Rs. 529. A and C are supposed to finish 19/23 of the work together. How much shall be paid to B ?

Sol: Work done by B = 1 - (19/23) = 4/23
(A + C) : B = (19/23) : (4/23) = 19:4
Bs share = RS.529*(4/23) = 92

41. A can do a certain job in 12 days. B is 60% more efficient than A. The number of days, it takes B to do the same piece of work, is.

Ratio of times taken by A and B = 160:100 = 8:5
8 : 5 :: 12 : x or 8x = 5*12 or x = 7 and half days.

42. A and B together can complete a piece of work in 35 days while A alone can complete the same work in 60 days. B alone will be able to complete the same work in: 

Sol:(A+B)s 1 days work = 1/35 and As 1 days work = 1/60
Bs 1 days work =(1/35 - 1/60) = 5/420 = 1/84 

43.A man can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. In what time can the son do it alone?

Son's 1 day's work
= (1/3-1/5)
=2/15.
The son alone can do the work in
= 15/2
=7x1/2 days.

44.Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by tanya to do the same piece of work is

Ratio of times taken by Sakshi and Tanya=125:100

= 5 : 4.

Suppose Tanya takes x days to do the work.

5 : 4 ::20 : x

x<=> (4×20/5)

<=> 16 days.


45.Kim can do a work in 3 days while David can do the same work in 2 days. Both of them finish the work together and get Rs.150. What is the share of Kim?

Kim's wages : David's wages
= Kim's 1 day's work : David's 1 day's work
= 1/3 : 1/2
= 2 : 3
Kim's share
= Rs.(2/5×150)
=Rs.60.

Courtesy : http://akulapraveen.blogspot.in/
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