In this section, we will begin with the Time and Work topic of your exams.  

The concept of Time and Work is moreover based on Unitary Method and LCM. Before we begin, we should know some basic relations:

1. Time(T) ∝ Work(W) i.e. To do more work we will need more time and vice versa.
2. Time(T) ∝ ¹/_Person(P) i.e. More person will do the same work in less time and vice versa.
3. Work(W) ∝ Person(P) i.e. To do more work we will need more person and vice versa.

These relations will help us to understand the concepts easily.

• Rule 1: If a person can do a piece of work in n days/hours/minutes then the work done in 1 day/hour/minute= 1/n and vice versa i.e. if the work done in 1 day/hour/minute is 1/n then the work will be done in n days/hours/minutes.  [V. Imp. Concept]

Note: Time taken to complete the work= Reciprocal of work done in 1 day or hour or minute.

Let’s take an example from to understand this rule.

• E.g.: Working together, P, Q and R reap a field in 6 days. If P can do it alone in 10 days and Q in 24 days, in how many days will R alone be able to reap the field.

P, Q and R reap a field in 6 days.

∴ Work done by them in one day= 1/n = 1/6

Similarly, work done by P and Q in one day is 1/10 and 1/24 respectively.

Let R alone can reap a field in x days.

Therefore, Work done by R in one day= 1/x

A/q,

1/10 + 1/24 + 1/x = 1/6

⇒ (1 × 12 + 1 × 5)/120 + 1/x = 1/6

⇒ 17/120 +1/x = 1/6

⇒ 1/x = 1/6 – 17/120

⇒ 1/x = (1 × 20 – 17 × 1)/120

⇒ 1/x = 20 – 17/120 = 3/120 = 1/40            

⇒ 1/x=1/40

⇒ x=40

Hence, R can do the work alone in 40 days (Ans).

• Rule 2: If a person is n times efficient than another person. Then,

1. Work done by first person: Work done by second person= n:1
2. Time taken (Efficiency) by first person: Time taken (Efficiency) by second person= 1:n

Let’s take an example to understand it,

• E.g.: Atul can do a piece of work in 18 days, Aravya is 50% more efficient than Atul. How much more time does Atul take than Aaravya to do the same piece of work?

Here, Aaravya is 50% more efficient than Atul,

∴ Work done by Aaravya: Work done by Atul= 1.5 : 1 or 3:2              (50% more efficient)

∴ Time taken(Efficiency) of Aaravya: Time taken(Efficiency) of Atul= 2:3

We have no. of days required by Atul= 18

So to balance the ratio, we have to multiply on both sides by 6.

Therefore, no. of days required by Aaravya= 12

 Hence, No. of days for Atul is 6 more than that of Aaravya (Ans).

[Time and Work Problems]

Now that we have discussed the basics let’s move ahead and see specific types and questions on them that have come previously.

• If A can do a work in m days and B can do the same work in n days, then-

Using Rule 1,

Work done by A and B in one day= 1/m + 1/n = (m + n)/mn

Time taken by A and B to complete the work= Reciprocal of work done in 1 day

                                                                         = mn/(m + n)

Let’s solve an example questions.

• E.g. 1: Aman and Ajay can build a wall in 9 days and 12 days respectively. In how many days can they finish the work if they work together?

By LCM method:

Step 1: Take LCM of the individual time taken.

L.C.M. of 9 and 12 = 36

Step 2:  Suppose LCM is the work done.

Let the work done= 36 units

Step 3: Use Unitary Method

Aman finished 36 units work in 9 days

∴ Work done in 1 day= 36/9 units = 4 units

Ajay finished 36 units work in 12 days

∴ Work done in 1 day= 36/12= 3units

Total work done in 1 day by Aman and Ajay together = 4 + 3 = 7 units

∴ Required time if they work together = Total work/Work done in 1 day

                                                                             =36/7 = 5¹/₇ days (Ans).

By direct formula:

Total work done by Aman and Ajay= mn/(m + n)

  = 9 × 12/(9 + 12) days

= 108/21days         [Dividing numerator and denominator by 3]

 = 36/7 days = 5¹/₇ days (Ans).

Note: This formula is limited for two people only. If we have three people in the problem then we will go by the LCM Method or use the Rule 1 that we discussed earlier.

• E.g. 2: Ravi, Rohan and Rajesh alone can complete a work in 10, 12 and 15 days respectively. In how many days can the work be completed, if all three work together?

Using LCM Method:

LCM of 10, 12 and 15= 60

Let the total work done= 60 units

Work done by Ravi in one day = 60/10 = 6units

Work done by Rohan in one day = 60/12 = 5units

Work done by Rajesh in one day = 60/15 = 4units

∴ Total work done by them in 1 day= 6 + 5 + 4=15 units

Hence, Total time taken to complete the work= 60/15= 4 days (Ans).

Using Rule 1:

Step 1: Take the reciprocal of individual time taken to do the work.

Ravi, Rohan and Rajesh alone can complete a work in 10, 12 and 15 days respectively.

Work done in 1 day by Ravi, Rohan and Rajesh = 1/10, 1/12 and 1/15 respectively [Using Rule 1]

Step 2: Add the reciprocals to get the work done together in one day

Work done by them in one day if they work together= 1/10 + 1/12 + 1/15

 = (1 × 6 + 1 × 5 + 1 × 4)/60

 = 6 + 5 + 4/60

 = 15/60.

Step 3: Take the reciprocal of the sum to get the total time taken

Time taken to complete the work= Reciprocal of work done in 1 day or hour or minute.

∴ Total time taken= Reciprocal of 15/60 = 60/15 = 4 days (Ans).

• If M₁ persons can do W₁ work in D₁ days working in T₁ hours in a day and M₂ persons can do W₂ work in D₂ days working T₂ hours in a day, then the relationship between them is:

M₁ × D₁ × T₁/W₁ = M₂ × D₂ × T₂/ W₂

• E.g. 1: 10 people can do a work in 30 days. In how many days, can 15 people complete double the work?

Using Unitary Method:

Here, 10 people can do a work in 30 days

∴ 1 person can do the work = 30 × 10 days

Hence,

15 people can do the work = ^{30 × 10}/₁₅

                                                = ^{30 × 10}/₁₅

                                                = ^{2 × 10}/₁

                                                = 2 × 10 days

                                                = 20 days

Same work was done by 15 people in 20 days.

∴ Double the work will be done by them in 20×2 days = 40 days (Ans).

Using Formula,

Here we have days, number of men and work given in the question so we will pit all the known quantities.

Let the work done by 10 people be W.

Since 15 people have to complete double the work

Then, work to be done by 15 people = 2 × W

Putting in formula,

M₁ × D₁ × T₁/W₁ = M₂ × D₂ × T₂/ W₂

⇒ ^{10× 30}/_W = ^{15× D2}/ _2×W

⇒ ^{10 × 30}/_W = ^{15 × D2}/ _{2 × W}

⇒ ³⁰⁰/₁ = ^{15 × D2}/ _{2 × 1}

⇒ 300 = ^{15 × D2}/ ₂

⇒ 300 × 2= 15 × D₂

⇒ D₂ = 300 × 2/15

         =20 × 2 days = 40 days (Ans).

(This solution also looks too long but we will avoid unnecessary writing and put everything directly in the formula)

• E.g.2: 30 men working 5 hours a day can do a task in 16 days. In how many days will 40 men working 6 hours a day do the same task?

M₁= 30, T₁= 5 h, D₁= 16 days

M₂= 40, T₂= 6 h, D₂=?

Suppose the work done by them is W

M₁ × D₁ × T₁/W₁ = M₂ × D₂ × T₂/ W₂

⇒ 30 × 16 × 5/W = 40 × D₂ × 6/ W

⇒ 30 × 16 × 5 = 40 × D₂ × 6

⇒ ^{30 × 16 × 5}/_40 × 6 = D₂

⇒ ^{5 × 16 × 1}/_{8 × 1} = D₂

⇒ D₂ = ^{5 × 2}/ ₁ = 5 × 2= 10 days (Ans).

• E.g.3: Five women can do a piece of work in thirty six days. If the ratio between the capacity of a man and a woman is 3:1, then find how many days it will take 5 men to complete the same work?

We have, capacity of a man and a woman = 3:1

⇒ M:W=3:1 or M=3, W=1

This means work done by 3Women= work done by 1 man

In this type of question while putting the values in the formula we will have to put the values of M and W from the given ratio.

Putting everything in formula,

M₁ × D₁ = M₂ × D₂

⇒ 5W × 36 = 5M × D₂

⇒ 5 × 1 × 36 = 5 × 3 × D₂

⇒ 5 × 36 = 15 × D₂

⇒ D₂= 5 × 36/15

         = 36/3 = 12 days (Ans).

[Time and Incomplete Work]

The questions with each we dealt with in our previous Blog are of certain types in which work was completed.

But now we will deal with some problems in which work is incomplete or someone backed out from the work after a certain time.

• E.g. 1: Smita can finish a work in 12 days and Sam can finish the work in 9 days. After working together for 4 days they both leave the job. What is the fraction of unfinished work?

Here, we have a problem in which the work is not complete. Let’s see how to solve it.

• This type of problem has no specific formula. So, whenever we see that the work is not complete we will try to solve these questions with either Unitary Method or Rule 1.

Let’s solve the above problem using both the methods.

Using LCM Method:

LCM of 12 and 9= 36

Let the total work to be done =36units

∴ Work done by Smita and Sam in one day = 36/12 i.e. 3 units + 36/9 i.e. 4 units = 7 units.

Both of them have worked for 4 days,

∴ Work done in 4 days= 4 × Work done in one day = 4 ×7 units = 28 units.

Hence, fraction of unfinished work= (36 – 28)/36 = 8/36 = 2/9 (Ans).

Using Rule 1:

∵ Smita can finish a work in 12 days and Sam can finish the work in 9 days.

∴ Work done by them in 1 day = 1/12 + 1/9                                           [Using Rule 1]

                                                       = (1 × 3 + 1 × 4)/36 = 7/36 units

Work done in 4 days = 4 × 7/36 units = 28/36 units

Unfinished work = (36 – 28)/36 = 8/36 = 2/9 (Ans).

Note: Both the methods are feasible and not very time consuming if we have command on LCM and addition of fractions. So, we can choose any of the method according to our convenience.

[Wages]

Let’s see an example to understand the wages questions.

• E.g.: A, B and C are employed to complete a work in Rs. 784. If A and B together complete 23/28 of the work. Calculate the wages (in Rs.) paid to C?

Work done by A and B = 23/28

Therefore, Work done by C = (28 – 23)/28 = 5/28

Hence, share of C = Work done × Wages

                                 = ⁵/₂₈ × 784

                                 = ⁵/₂₈ × 784

                                 = 5× 28 = Rs. 140 (Ans).

We can see that the questions on wages are just an extended part of Time and work normal questions. All we need is to see the share of each individual in it. Question asked from wages part is very few and they are easy to solve also.

• If A, B and C can do a piece of work in a, b and c days respectively, the ratio of their wages will be-        A:B:C: = bc:ac:ab and share of wages will be-

Share of A= {bc/(bc + ac + ab)} × Total wages

Share of B= {ac/(bc + ac + ab)} × Total wages

Share of C= {ab/(bc + ac + ab)} × Total wages

• E.g.: If A, B and C can complete the work in 4, 5 and 6 hours respectively. If they all work together and receive Rs. 777 as wages, then find the share of A.

Using LCM Method:

LCM of 4, 5 and 6 = 60

Let the total work be 60 units.

Work done by A, B and C in 1 hour = 60/4, 60/5 and 60/6 respectively

                                                           = 15, 12 and 10 respectively

Total work done in 1 hour = 37 units

Total time taken = 60/37 hours

Hence, Work done by A in 60/37 hours = 15 × 60/37

Share of A = {Work done by him/ Total work} × Total wages

                   = {(15 × 60/37)/60} × 777

                  = {^{15 × 60}/_{37 × 60}} × 777

                  = {^{15 × 1}/_{37 × 1}} × 777

                   = ¹⁵/₃₇ × 777

                   =¹⁵/₁ × 21

                   =15 × 21 = Rs. 315 (Ans).

By formula:

Ratio of work done by A, B and C= bc:ac:ab

                                                = 5 × 6 : 4 × 6 : 4 × 5

                                                =30:24:20    [Converting to simplest form by dividing by CF 2]

                                                =15:12:10

∴ Share of A = {¹⁵/_{(15 + 12 + 10)}} ×777

                       = ¹⁵/₃₇ × 777

                       = 15 × 21 = Rs. 315 (Ans).

[Pipes and Cistern]

• A pipe can fill/empty a tank in ‘m’ hours and another pipe can fill or empty the same tank in ‘n’ hours. Then –

1. If both pipes fill or empty the tank then the time taken to fill or empty tank when both pipes are opened, is t = mn/(m + n)

• E.g.: A jug has two holes. The first hole alone makes the jug empty in 15 minutes and 2^nd hole alone makes the jug empty in 20 minutes. If water leaks out at a certain rate, how long in minutes does it take to if both the holes together empty the jug.

By LCM Method:

LCM of 15 and 20 = 60

Let volume of the jug is 60 unit³.

∴ Volume emptied by first hole in 1 minute = 60/15 = 4 unit³

And, Volume emptied by second hole in 1 minute = 60/20 = 3 unit³

Volume emptied in 1 minute when both are open = 4 + 3 = 7 unit³

Hence, time taken to empty the jug = Volume of the jug/ Volume emptied by both hole in 1 minute = 60/7 minutes = 8⁴/₇ minutes (Ans).

By formula:

Time taken to empty the jug, t = mn/(m + n)

                                                         = 15 × 20/(15 + 20)

                                                         = ^{15 × 20}/₃₅

                                                         = ^{3 × 20}/₇ = 60/7 minutes

                                                         = 8⁴/₇ minutes (Ans).

It’s clear that formula method is very quick and short.

2. If first pipe fills the tank and second pipe empties the tank, then the time taken to fill the tank when both pipes are opened, is t = mn/(m – n), where m>n

Note: Cases in which one pipe fills and other empties: Take the value of ‘n’ from one which fills and value of ‘m’ from one which empties.

• E.g.: A tap can fill a tank in 50 minutes. If the tank has a leakage which alone is capable of emptying the tank in 2½ hours, the tank will now be filled in ______.

Convert minutes into hours,

50 minutes= 5/6 h and 2½ h = 5/2 h

By formula,

Time taken to fill the tank, t= mn/(m – n)

                                              = {⁵/₆ × ⁵/₂}/(⁵/₂ – ⁵/₆)

                                              = {⁵/₆ × ⁵/₂}/^{(5 × 3  – 5 × 1)}/₆

                                               = {⁵/₆ × ⁵/₂}/¹⁰/₆

                                               = ⁵/₆ × ⁵/₂ × ⁶/₁₀

                                               = ¹/₁ × ⁵/₂ × ¹/₂ = ⁵/₄ hours = 1 ¼ h = 1 hours (¼ × 60) minutes

                                                                                                   = 1 h 15 min (Ans).

3. If first pipe fills the tank and the second pipe empties the tank, then time taken to empty the tank when both pipes are opened, is t= mn/(n-m), where n>m
4. E.g.:  A tap can be filled in 90 minutes. There is a leakage which can empty the tank in 30 minutes. Therefore, the tank will be emptied in. By formula,

Time taken to empty the tank, t= mn/ (n – m)

                                              = 90 × 30/ (90 – 30)

                                               = ^{90 × 30}/₆₀

                                               = 90/2 minutes = 45 minutes (Ans).

• If three pipes can fill a tank separately in m, n and p respectively, then part of tank filled in 1 h by all the three pipes = (1/m + 1/n + 1/p) and

Total time taken to fill the tank =  mnp/ (np + mp + mn)

Note: If any pipe empties the tank then time taken by that pipe will be negative.

• E.g.:  A tank can be filled by two taps X and Y in 5 hrs and 10 hrs respectively while another tap Z empties the tank in 20 hrs. In how many hours will the tank be filled if all the three taps are kept open.

By LCM Method:

LCM of 5, 10 and 20 = 20

Let the volume of the tank be 20 unit³

Tank filled by X in 1 hour = 20/5 = 4 unit³

Tank filled by Y in 1 hour = 20/10 = 2 unit³

Tank emptied by Z in 1 hour = 20/20 = 1 unit³

Tank filled in 1 hour when all the three pipes are open= 4 + 2 – 1 = 5 unit³

Time taken to fill the complete tank = Volume of the tank/ Tank filled in 1 hour

                                                            = 20/5 hours = 4 hours (Ans).

By formula:

Here, m=5 h,

n= 10 h,

p= 20 h

Since, Z empties the tank so its time will be negative

Total time taken to fill the tank = mnp/(np + mp – mn)

                                                    =5 × 10 × 20/{10 × 20 + 5 × 20 – 5 × 10}

                                                    =5 × 200/{200 + 100 – 50}

                                                    =1000/250 = 4 hours (Ans).