Time and Work

M. Velrajan 

A. Suppose A can do a work in a days and B can do it in b days.

Then A and B together complete the work in ab / (a + b) days. 

For,  In one day A and B together complete 1/a + 1/b = (a + b) / ab of the work.

Hence A and B together complete the work in ab / (a + b) days.

B. Suppose A, B and C can do a piece of work in a, b and c days respectively. 

Then A, B and C together complete the work in abc / (ab + bc + ca) days.

For,  In one day A, B and C together complete 

1/a + 1/b + 1/c = (ab + bc + ca) / abc of the work.

Hence A, B and C together complete the work in abc / (ab + bc + ca) days.

B. 1. Suppose A, B and C can do a piece of work in a, b and c days respectively. Then we can find the days A, B and C together to complete the work as follows :  

In one day A, B and C together complete 1/a + 1/b + 1/c of the work.

Suppose m = lcm{a, b, c} and m = a × x = b × y = c × z.

Then 1/a + 1/b + 1/c = (x + y + z)/m

In one day A, B and C together complete (x + y + z)/m of the work

Hence A, B and C together complete the work in m/(x + y + z) days.

So, if A, B and C can do a piece of work in a, b and c days respectively

then A, B and C together to complete the work in  m/(m/a + m/b + m/c) days.

C. Suppose A can do a work in a days and B can do it in b days.

Suppose only A works for d days and leaves the work then B alone can complete the

work in b(a - d)/a days.

For,   The work completed by A in d days = d/a of the work

Remaining work = 1- d/a = (a - d)/a

B do 1/b of the work in one day.

Number of days required for B to complete the remaining work 

              = (1/(1/b)) × (a - d)/a = b(a - d)/a

This can be remembered as given below : 

                          Do in days         Remaining 

A started        :          a                  a - d

B completed   :          b                   x

Cross multiply and equate

                ax = b(a - d)

Also in a, b, d, x if any three are given then the remaining one can be obtained

from this relation.

D. Suppose A can do m times as B can do a work and B can do it in b days.

Suppose it is given that A and B together complete the work in d days.

Then b = md + d and A can do the work in a = b/m = d + d/m days. 

For,  In one day B completes 1/b of the work.

Hence in one day A completes m/b of the work and 

A can do the work in a = b/m days .

And in one day A and B together complete m/b + 1/b = (m + 1) / b of the work.

Hence A and B together complete the work in b / (m + 1) days.

Hence b / (m + 1) = d

i.e. b = (m + 1)d = md + d

And a = b/m = d + d/m

1. It takes 10 men to finish a work in 2 days. Four (4) men started to execute the

work but after one day 3 of them left. How many days will it take the lone man to

finish the remaining work?     

1. 12 days         2. 14 days         3. 15 days       4. 16 days 

Number of men required to finish the work = 2 × 10 = 20

4 men worked for 1 day

Men required for the remaining work = 20 - 4 = 16

It will take 16 days for the lone man to finish the work.

2. Two persons X and Y can do a piece of work in n days. If X alone can complete

the work in (n + 3) days and Y alone can do the work in (n + 12) days,

find the value of n.  1. 6          2. 10           3. 7           4. 5

n = number of days for X and Y together to complete the work 

   = (n + 3)(n + 12) / (n +3 + n + 12)   (By A.)

n(2n + 15) = n2 + 15n + 36 

n2 = 36 and n = 6

3. A can do ⅓ of a work in 4 days and B can do ¾  of the work in 15 days.

Together they will complete the work in

1. 4 ¼ days           2. 5 ¾ days         

3. 8 ¼ days            4. 7 ½ days 

Number of days for A = 4 × 3 = 12

Number of days for B = 15 × (4/3) = 20

Number of days for A and B = (12 × 20)/(12 + 20)   (By A.) 

                                                  = (12 × 20)/32 = 15/2 = 7 ½ 

4. Two persons A and B can do a piece of work in 45 and 40 days respectively.

They started their work together but A left the work after some days and the

remaining work was completed by B in 23 days. Find after how many days did 

A leave?      1. 11    2. 10    3. 8    4. 9

By A., A and B together do in (45 × 40) / (45 + 40) 

                                                              = (45 × 40) / 85 = 360/17 days

Suppose A left after x days.

By C.,

                              Do in days         Remaining

A and B starts  :    360/17             (360/17) - x

   B  completes  :     40                          23

Cross multiply and equate

(360/17)23 = ((360/17) - x)40

360 × 23 = (360 - 17x)40

17 × 40 x = 360(40-23) = 360 × 17

x = 9     Ans. 4

5. 7 men can complete a work in 52 days. In how many days will 13 men finish the same

work? (a)20 days          (b)13 days            (c)7 days          (d)28 days

14 men can complete in 52/2 = 26 days.

Since the days for 13 men have to be more than that for 14 men, Ans. 28 days.

Aliter : One man complete  1/(7 × 52) of work in a day.

13 men complete (1/(7 × 52)) × 13 =  1/28 of work in a day.

Ans. 28 days.

6. Seven men working 9 hours a day can do a piece of work in 30 days. 

In how many days will 10 men working for 7 hours a day do the same work?   

(a)28 days   (b)30 days    (c)32 days       (d)27 days

Total man hour for the work = 7 × 9 × 30

Hence 10 × 7 × ? = 7 × 9 × 30 and hence  ? = 9 × 3 = 27

7. A can do a certain job in 12 days. B is 60% more efficient than A. How many

days does B alone take to do the same job?       

(a)6 days         (b)7 ½ days            (c)8 days              (d)8 ½ days

B is 60% more efficient than A. 

B can do 160/100 times as A can do. 

Since A alone do in 12 days, By interchanging A and B in the result D.,

B alone do in 12/(160/100) = 12 ×  100/160 = 15/2 days

Aliter : 

Part of work by B in a day = (1/12) × (160/100) = 2/15 

Ans. 15/2 days

8.  A can complete ⅔ part of a work in 10 days. A can complete ⅓ part of the

same work in 

(a)3 days                   (b)4 days                   (c)5 days                   (d)6 days

⅓  = ½ of ⅔ 

Ans.  ½ × 10 = 5

9. First pipe can fill a tank in 12 hours. Second pipe can fill the same tank in 

6 hours . Third pipe in 4 hours. How long will it take to fill  the tank if 

all the 3 pipes are opened simultaneously?

(a)2 hrs.                   (b)3hrs                          (c)4hrs                    (d)12hrs

LCM of 12, 6, 4 is 12, 12 = 12 × 1 = 6 × 2 = 4 × 3 and 1 + 2 + 3 = 6. 

By B.1., the tank becomes full in 12/6 = 2 hrs.

Aliter: 1/12 + 1/6 + 1/4 = (1 + 2 + 3)/12 = 1/2 . Ans. 2 hrs. 

Note that the calculation in the first method is the same as the second but

without writing the reciprocals. We follow the first one.   

10. Two taps can fill a tank in 30 minutes and 40 minutes. Another tap can empty it in

24 min. If the tank is empty and all the three taps are kept open, in how much 

time the tank will be filled.

(a) 1 ½ hours            (b)two hours                     (c)one hour           (d) 2 ½ hours

LCM(30, 40, 24) = 120

120 = 30 × 4 = 40 × 3 = 24 × 5 and 4 + 3 - 5 = 2 (since the 3rd pipe empty the

tank, we have to take -5 instead of 5)

By B.1., Ans. 120/2 = 60 minute = 1 hr

Aliter:    In 1 minute = 1/30 + 1/40 - 1/24 = (4 + 3 - 5)/120 

                                 = 2/120 = 1/60 of the tank is filled up.

      Ans. One hour

11. A tap can  fill a tank in 15 minutes. Another tap can empty it in 20 minutes.

Initially the tank is empty, if both the taps start functioning at the same time, 

when will the tank become full?

(a)1 hour        (b)3 hours                   (c)2 hours                (d)4 hours

LCM(15, 20) = 60 and 60 = 15 × 4 = 20 × 3.

4 - 3 = 1 (since the 2nd pipe empty the tank, we have to take -3 instead of 3)

By B.1., the tank becomes full in 60/1= 60 minutes = 1 hr.

12. A, B and C together can finish a piece of work in 4 days. A alone can do it in

12 days and B alone in 18 days. How many days will be taken by C to do it alone?

(a)10 days                    (b)12 days                         (c)9 days                  (d)18 days

C = (A+B+C) - A - B

LCM of 4, 12, 18 = 36. 

36 = 4 × 9 = 12 × 3 = 18 ×2

9 - 3 - 2 = 4 (since C = (A+B+C) - A - B, -3 and -2 is taken instead of 3 and 2)

By B.1., C alone do the work in  36/4 = 9 days  

Aliter :  Suppose C alone can do the work in c days.

Then in 4 days A, B and C together do 4(1/12) + 4(1/18) + 4(1/c) of the work.

Since A, B and C together completes the work,

4(1/12) + 4(1/18) + 4(1/c) = 1

(1/12) + (1/18) + (1/c) = ¼ 

1/c = ¼ - 1/12 - 1/18 = (9 - 3 - 2) / 36 = 4/36 = 1/9

c = 9 days  

13. If A and B together complete a work in 20 days. If A alone completes the work

in 24 days, then B alone completes the work in   

(a)14 days             (b)44 days          (c)120 days            (d)48 days

B = (A + B) - A

LCM(20, 24) = 120

120 = 20 × 6 = 24 × 5

6 - 5 = 1 (since B = (A + B) - A, we take -5 instead of 5)

By B.1., B alone do the work in 120/1 = 120 days

14. A and B can do a piece of work in 10 days; B and C in 15 days; C and A in 18

days. In how many days can B alone do it?       

(a)30 days            (b)20 days               (c)12 days                (d)18 days

(A + B) + (B + C) - (C + A) = 2B

LCM(10, 15, 18) = 90

90 = 10 × 9 = 15 × 6 = 18 × 5

9 + 6 - 5 = 10 (since (A + B) + (B + C) - (C + A) = 2B, we take -5 instead of 5) 

By B.1., 2B can do the work in 90/10 = 9 days.

Hence B alone do the work in 2 × 9 = 18 days   

15. P, Q and R together do a piece of work for Rs.535. P working alone can do it in

5 days. Q alone can do it in 6 days  and R alone can do it in 7 days. Then what will

be the share of R for his work?   

(1) Rs.100 (2) Rs.150 (3) Rs. 200 (4) Rs. 250

The ratio of work done by P, Q and R is 15 : 16 : 17 

i.e. 42 : 35 : 30

42 + 35 + 30 = 107

For 107 amount = 535, 

hence for 30 amount = (535/107) × 30 = 5 × 30 = 150

16. If 12 compositors can compose 60 pages of a book in 5 hours, how many

compositors will compose 200 pages of the book in 20 hours?         

(a)8           (b)10            (c)12           (d)11

In one hour a compositor compose 60/(12 × 5) = 1 page. 

In 20 hours a compositor composes  1 × 20 = 20 pages.

Number of compositers to compose 200 pages in 20 hrs. = 200/20 = 10. 

17. A can do a work in 12 days; B in 6 days and C in 3 days. A and B start working

together and after a day, C joins them. The total number of days required to complete

the work is 

 (a)2 2/7  days            (b)1 2/7  days             (c)2 1/7  days          (d) 1 1/7  days  

A + B can do it in (12 × 6) / (12 + 6) = 4 days

A + B + C = (A + B ) + C can do it in (4 × 3) / (4 + 3) = 12/7 days

By C.,

A + B started              :   4       4 - 1 = 3

A + B + C completed : 12/7            x

Cross multiply and equate

4x = 36/7

  x = 9/7 

Total number of days = 1 + 9/7 = 16/7 = 2 2/7

Aliter :

LCM (12, 6, 3) = 12

12 = 12 × 1 = 6 × 2 = 3 × 4

By B.1., A, B and C do the work in 12/(1+2+4) = 12/7 days

In one day A and B can do (1+2)/12 = 1/4 of work

A, B and C do the remaining ¾ of work in = ¾ × (12/7) = 9/7 days

Total number of days = 1 + 9/7 = 16/7 = 2 2/7 days

18. If 15 men or 24 women or 36 boys can do a piece of work in 12 days, working

8 hours a day, how many men must be associated with 12 women and 6 boys to do

another piece of work 2.25 times as great in 30 days working 6 hours a day? 

(a) 24               (b) 18        (c) 10 (d) 8

Since 15 men by working 8 hours a day do the work in 12 days,

One man do in one hour 1/(15 × 12 × 8) of the work

One man do in 30 × 6 hours (30 × 6) / (15 × 12 × 8) of the work.

Similarly,

One woman do in 30 × 6 hours (30 × 6) / (24 × 12 × 8) of the work

One boy do in 30 × 6 hours (30 × 6) / (36 × 12 × 8) of the work 

Let x be the number of men to be associated with another work.

x man, 12 women and 6 boys do in 30 × 6 hours 

[x(30 × 6) / (15 × 12 × 8)] + [12 × (30 × 6) / (24 × 12 × 8)] 

              + [6 × (30 × 6) / (36 × 12 × 8)] of the work

= (x/15 + 12/24 + 6/36)×[(30 × 6) / (12 × 8)] of the work

Hence

(x/15 + 12/24 + 6/36) × [(30 × 6) / (12 × 8)] = 2.25 = 9/4

x/15 + ½ + ⅙ = (9/4) × [(12 × 8) / (30 × 6)] = (9 × 12 × 8) / (4 × 30 × 6) = 6/5

(2x + 15 + 5) / 30 = 6/5

2x + 20 = 36

2x = 16

x = 8

By duplicating the steps of the problem 18, we can drive a formula for such problems as given below:

E. Suppose there are three categories people A, B and C.

Suppose ‘a’ number of category A or b of category B or c of C can do a work in p days, working h hours a day. And x of category A along with y of category B and z of C can do another work m times of the first work in q days, working k hours a day. Then

x/a + y/b + z/c = m × (p × h) / (q × k)

= m × (Total time for first work / Total time for second work)

(Out of the 11 values in the above if any 10 values are given then we can find the remaining one using the above)

For, Since a of A by working h hours a day do the work in p days,

One of A do in one hour 1/(a × p × h) of the work

Similarly,

One of B do in one hour 1 / (b × p × h) of the work

One of C do in 1 hour 1 / (c × p × h) of the work.

Hence

x of A, y of B and z of C do in 1 hour

[x / (a × p × h)] + [y / (b × p × h)]

+ [z / (c × p × h)] of the work

= (x/a + y/b + z/c)×[1 / (p × h)] of the work

And x of A, y of B and z of C do in q × k hours

= (x/a + y/b + z/c)×[(q × k) / (p × h)] of the work

Since the another work is m times of the work,

(x/a + y/b + z/c)×[(q × k) / (p × h)] = m

i.e. x/a + y/b + z/c = m × (p × h) / (q × k)

= m × (Total time for first work / Total time for second work)

Aliter for 18.: We consider man as category A, woman as B and boys as C.

We are given a = 15, b = 24, c = 36, p = 12, h = 8, q = 30, k = 6, y = 12, z = 6 and

m = 2.25 = 9/4.

We have to find x.

By E., x/a + y/b + z/c = m × (p × h) / (q × k)

x/15 + 12/24 + 6/36 = (9/4) × [(12 × 8) / (30 × 6)] = (9 × 12 × 8) / (4 × 30 × 6) = 6/5

x/15 + ½ + ⅙ = 6/5

(2x + 15 + 5) / 30 = 6/5

2x + 20 = 36

2x = 16

x = 8