# Time & Work Concepts - Sibanand Pattnaik

• I will explain some important question types from Time & Work topic. We will learn via examples so that you can relate to the concept easily.

Questions involving 2 persons

There are various ways of doing such questions. Lets discuss couple of them

Ques 1: A can do a work in 10 days and B can do it in 15 days. In how many ways will they do the work together?
Ques 2: A can do a work in 24 days and with B can do it in 6 days. In how many days can B alone do the work?

UNITARY METHOD:

Ans 1:
Work done by A in 1 day= 1/10
Work done by B in 1 day = 1/15
Work done by both in 1 day = 1/10 + 1/15 = 10/60
Therefore time taken by them to dp the work together = 60/10 -= 6 days

Ans 2:
Work done by A in 1 day = 1/24
Work done by A and B together in 1 day = 1/6
Therefore work done by B in 1 day = 1/6 – 1/24 = 1/8

LCM METHOD (preferred one)

Ans 1:
Lets assume the total work as LCM of time taken by A and B individually = LCM (10, 15) = 60 units
Total work = 60
Time taken by A = 10 days
So work done by A in 1 day = 6 units
Time taken by B = 15 days
So work done by B in 1 day = 4 units
So work done by A and B in 1 day = 10 units
So time taken by A and B to complete total work i.e. 60 units = 60/10 = 6 days

Ans 2:
Lets assume the total work as LCM of time taken by A and B individually = LCM (6, 24) = 24 units
Total work = 24
Time taken by A = 24 days
So work done by A in 1 day = 1 unit
Time taken by A and B together = 6 days
So work done by B in 1 day = 24/6 = 4 units
So work done by B in 1 day = 4 – 1 = 3 units
So time taken by B to complete total work i.e. 24 units =24/3 = 8 days

Questions involving 3 persons

Q1) . Time taken by A, B and C to complete a work alone is 4, 6 and 12 days respectively. In how much time will they complete the work, working together?

Let us assume the total work to be LCM (4,6,12) = 12 units
So work done 1 in day by:
A = 12/4 = 3 units
B = 12/6 = 2 units
C = 12/12 = 1 units
So work done by all 3 together in 1 day = 3 + 2 + 1 – 6 units
So time taken to complete 12 units = 12/6 = 2 days
Q2. Time taken by A and C to complete a work alone is 10 and 15 days respectively. They complete the work, working together with B in 4 days. In how much time can B alone complete the work?
Let us assume the total work to be LCM (10,15,4) = 60 units
So work done 1 in day by:
A = 60/10 = 6 units
C = 60/15 = 4 units
A+B+C = 60/4 = 15 units
So work done by B in 1 day = 15 – 6 - 4 = 5 units
Time taken by B to complete the whole work together = 60/5 = 12 units

Q)
A and B complete a work in 10 days. A and C complete the same work in 12 days. B and C do that work in 15 days.
a) In how much time all 3 complete the work?
b) In how much time will each one individually do the work?

Let the total work = LCM (10, 12, 15) = 60
Therefore work done by A + B = 60/10 = 6 units
Therefore work done by A + C = 60/12 = 5 units
Therefore work done by B + C = 60/15 = 4 units
Therefore total work done by 2(A+B+C) in 1 day= 15 units
Therefore total work done by (A+B+C) in 1 day = 7.5 units

a) Therefore time taken by (A+B+C) = 60/7.5 = 8 days

b) To find the time taken by C alone
Work done by A+B+C in 1 day = 7.5
Work done by A+B in 1 day = 6 units
Therefore work done by C alone in a day = 1.5 units
Therefore time taken by C alone = 60/1.5 = 40 days
Similarly time taken by B alone = 24 days
Time taken by A alone = 120/7 days

Joining and Quitting

Many of the challenging question in Time and Work concern Joining and Quitting situation. In such questions workers either join or leave midday

Example1:

A and B start a work which they can complete in 10 days. B left 2 days before the scheduled completion due to which it took 1 extra day to complete the work. In how much time can A and B complete the work working alone?

Since the total time taken by them is 10 days, therefore in 2 days they must complete 2/10 = 20% of the work
Alone took 1 extra day so he took 2 + 1 = 3 days to complete the work
Therefore A takes 3 days to complete 20% of the work
Therefore A can complete the whole work in (100/20) x 3 = 15 days
Using methods discussed earlier we can calculate that B can complete the whole work in 30 days

Example 2:

A starts a work which he can complete in 12 days. B joins 2 days later and they complete the work in the next 2 days. In how many days can B alone complete the work?

A has worked for 2 + 2 = 4 days
So work done by A = 4/12 = 1/3rd of the total work
So work done by B = 2/3rd of the total work
Time taken by B to complete 2/3rd of the total work = 2 days
So time taken by B to complete the whole work = 2 x 3/2 = 3 days

Example 3:

A and B start a work. A can complete the work working alone in 10 days and B can complete it in 15 days. After 3 days C joins them. The work is completed in 1 day. In how much time can C alone complete the work?

Let the total work be LCM (10, 15) = 30 units
So work done by A in 1 day = 30/10= 3 units
Work done by B in 1 day = 30/15= 2 units
A+B worked for 3+1 = 4 days
Work done by A in 4 days = 12 units
Work done by B in 4 days = 8 units
Work done by A+B in 4 days = 12 + 8 = 20 units
Work done by C in 1 day = 10 units
SO time taken by C to complete the whole work i.e.e 30 units = 30/10 = 3 days

Example 4:

A start a work. After 1 day, B joins. After another 2 days, C joins. Work is completed in the next 3 days. A alone could have completed the work in 12 days. Ratio of efficiency of B and C is 3:2. In how many days can B and C together complete the work?

Number of days for which A worked = 1 + 2 + 3 = 6 days
So work done by A = 6/12 = 1/2
Work done by B and C = 1/2
Now B worked for 5 days and C for 3 days
Ratio of efficiency is 3:2
Let u s assume that B does 3 units every day and C does 2 units every day
So total work done by B + C = 3 x 5 + 2 x 3 = 21 units
Total work 21 x 2 = 42 units (since B+C have done half of the total work)
So total time taken by B+C = 42/(3+2) = 8/4 days to complete the whole work

MDH = Man Days Hours

In such questions, one or more group of persons do a work. The efficiency of each person within a group is assumed to be same.

Formula is given as : (M x D x H)/W = (M1 x D1 x H1)/W1

5 men can complete a work in 10 days working 8 hours day. In how many days will 20 workers working 5 hours day complete twice the work?

(5x10x8) = (20x5xD)/2
Therefore D = 8 days

6 men and 3 women can complete a work working 8 hours for 12 days. In how much time will 12 women complete the same work working 6 hours every day, if the efficiency of each man is twice that of a woman?

1M= 2 W
5M = 12 W
As given
(6M+3W) x 8 x 12 = 12W x 6 X D
15W x 8 x 12 = 12W x 6 x D
OR D = 20

8 cows and 12 goats can give x litres of milk in 25 days. 6 cows and 11 goats can give the same quantity of milk in how many days, if 2 cows give as much milk as 3 goats?

((8c+12g)×25)/x=(6c+11g)D/x
Since 2c=3g, therefore
((12g+12g)×25)/x=(9g+11g)D/x
Therefore D= 30 days

One persons starts a work. Next day one more person joins him. Next day another person joins them. One person alone can complete the work in 110 to 130 days, the in how many days will the work be completed?

If we take one person as 1 unit, the the number of units done per day goes like a series: 1, 2, 3, 4…
Sum of the series will be : n(n+1)/2
This should lie between 110 and 130
The only possible value of n for which this is true is 15

In a fortress there is enough food for 100 persons for N days. 10 days later 100 more persons join. 10 days later, these 100 persons leave. As a result food is exhausted 5 days later. If these 100 persons had not left then the food would have lasted for how many days?

Total food usage = 100 x 25 + 100 x 10 = 3500
If the 100 persons would not have left then the number of days:
100 x 10 + 200 x D = 3500
Therefore D = 12.5 days
So total 22.5 days

In a ground grass grows at a uniform rate every day. If 7 cows are left to graze, then the whole ground will be bereft of grass in 12 days. If 14 cows are allowed to graze, then will eat up all the grass in 5 days. How many cows are required to completely mow the grass in 3 days?

Ans - 22

Negative Work

Negative work is similar to positive work except that one person is destroying the work. SO instead of the plus sign we will have a minus sign. That is the only difference.

If A can complete a work in 10 days and B can destroy a work in 20 days, then in how many days will the work be completed?
Let total work = 20 units
So work done by A in 1 day = 2 units
Work destroyed by B in 1 day = 1 unit
So work done in 1 day = 1 unit
Time taken = 20/1 = 20 days

A can do a work in 30 days, B can do it in x days and C can destroy the work in 50 days. If A and B work together, alternating with C, (i.e. A & B together on day 1 and C alone on day 2 and so on), they complete the work in 50 and 5/7 days how long will it take for B to complete the work alone?

A can do a work in 30 days, B can do it in x days and C can destroy the work in 50 days
150Pots (assume total work as 150 pots)
A 30 5p
B x YP
C -50 -3p
A & B together on day 1 and C alone on day 2 and so on, they complete the work in 50 and 5/7 days
Hence, A& B worked for 25+5/7 days and C for 25 days
Total number of pots made= (5+y)(25+5/7)+(-3)25=150 pots
Y = 15/4 pots
X = 150/(15/4) = 40 days

Pipes and Cisterns

It’s a classic case of negative work

A tap can fill a cistern in 12 hours. There is a leak in the bottom which can empty it in 20 hours. The cistern was empty when the tap was turned on. In how many hours will the cistern be filled?

Let the capacity of cistern be 60
Therefore work done by tap in 1 hour = 5
Work done by leak in 1 hour = 3
So net work done = 2
Time taken = 60/2 = 30

Slight Tricky question

There is a cistern of capacity 100 litres. A tap can fill the cistern in 10 hours. There is a leak at the half way mark of the cistern because of which the it took 15 hours to fill the cistern. Find the time in which the filled cistern can be emptied if there were 2 leaks at the half way mark?

Answer: Can never be emptied as both the leaks are at half way mark

There is a cistern which has 4 equidistant leaks with one being right at the bottom and one at 1/4th mark from the top. A tap is opened which can fill a cistern in 120 hours (without the leak). Ratio of efficiency of each leak to the tap is 1:6. In how much time will the tank be filled?

Lets divide the cistern in 4 parts each part will take 30 hours. The bottom part of cistern..only 1 leak (the bottom leak) is in play..so the efficiency will reduce from 6 to 5..so the time will increase from 5 to 6..so time taken = 6/5 * 30 = 36 hours.
Similarly in the area between 14th from bottom and half way mark, 2 leaks are applicable, so efficiency will go from 6 to 4, the time will go from 4 to 6, so the time taken will be 45 hours similarly for the next one, efficiency will go from 6 to 3, so time from 3 to 6 and therefore 60 hours and finally for the top part the time taken will be 90 hours.
So the total time taken will be 231 hours

Alternate days:

A can complete a work in 10 days and B in 12 days. They start working such that one person works on one day and the other person works on the other day. When will the work be completed?

In such questions, often the answer depends on who starts the work.
For example, let the total work be 60 units
So A can do 6 units in a day and B can do 5 units in a day
So in 2 days they will do 11 units
In 10 days they will do 11 x 5 = 55 units
Now if A works he will complete the work in 5/6 days and if B works he will complete the work in 1 day

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