Time and Work is one of the topics asked under the miscellaneous questions in the Quantitative Aptitude Section. It may also be combined with data sufficiency or wages based questions. We received many requests from our readers about tips to approach the miscellaneous section of Quantitative Aptitude. So, today we are going to provide you with some study tips with examples that should help you understand one of the topics of miscellaneous category better.
Before we move on to the types of problems that you may face in this topic, understand the following formulae 
 If A can do a piece of work in x days, then A’s 1day work = 1/x
 If A’s one day work is 1/x, A can finish the total work in x days.
 If A is x times as efficient as B then
The ratio of work done by A : B individually = x : 1
The ratio of time taken by A and B individually to finish the work = 1 : x
Now let us have a look at the types of problems that you may encounter under this topic
1. The first type of problem is the most basic one that may be asked  Calculation of time from work or vice  versa.
For instance 
A takes 8 days to finish a piece of work. B takes 10 days to finish the same work. How long will it take to finish the work when both of them are working together?
Now for such questions, A’s 1day work = ⅛
B’s 1day work = 1/10
Work done by both A and B together in one day = ⅛ + 1/10 = 9/40
Hence, A and B together will finish the work in 40/9 days.
Similarly, when n people are involved, you can follow the above approach. For instance,
B and C together can complete the work in 8 days. A and B together can complete the same work in 12 days and A and C together can complete it in 16 days. In how many days can A, B, C can complete the same work?
For this type of question, find the 1day work of A and B, B and C and A and C.
1day work of B and C = 1/8  (a)
1day work of A and B = 1/12 (b)
1day work of A and C = 1/16 (c)
Adding (a), (b) and (c),
1day work of B and C + 1day work of A and B + 1day work of A and C = 1/8 + 1/12 + 1/16
2 * (1 day work of A + B + C ) = 13/48
1 day work of A + B + C = (13/48) / 2 = 13/96
So, the number of days taken by A, B, C to finish the work = 96/13.
2. Another type of problem that you may face is when you need to normalise the time unit.
A can do a piece of work in 9 days by working 7 hours each day. B can do it in 7 days, working 6 hours per day. How long will it take them to complete the work together if both of them work for 504 minutes per day?
In this type of question, since the time unit mentioned is not the same for both, so you need to find per hour work of A and B.
A’s 1 hour work = 1/9*7 = 1/63
B’s 1 hour work = 1/7*6 = 1/42
Now, A and B’s one hour work = (1/63 + 1/42) = 5/126, i.e. they will finish the work in 126/5 hours. But as per question, they work 504 minutes per day i.e. 504/60 hours per day. So, number of days = (126/5) / (504/60) = 3 days.
3. Another type of problem that you may face is efficiency based.
For instance 
X can do a piece of work in 12 days. Y is 80% more efficient than X. How many days will Y take to complete the same work alone.
Now as per question Y does the work 1.8 times more efficiently than X (80% more).
So, ratio of time taken by X : Y = 1.8 : 1 i.e. 12/Y = 1.8/1
Y = 12/1.8 i.e. 20/3 days
4. Another case that you may face is when one person leaves the work midway.
For instance 
X can finish a piece of work in 18 days and Y can do the same work in 15 days. Y worked for 10 days and left the job. In how many days, X alone can finish the remaining work?
In this case, X’s 1day work = 1/18, Y’s 1day work = 1/15
Work done by Y in 10 days = 10 * 1/15 = ⅔
Remaining work = 1  ⅔ = ⅓
X finishes 1/18th work in 1 day, so he will finish ⅓ work in = (⅓) / (1/18) = 6 days
5. You may also encounter problems based on wages.
For instance
X and Y do a piece of work for Rs.600. X alone can do it in 6 days while Y alone can do it in 8 days. With the help of Z, they finish it in 3 days. Find the share of each.
Now, for this question, we’ll find one day work of each and the share of wage will be divided in the ratio of their one day work.
Z’s one day work = ⅓  (⅙ + ⅛) = 1/24
X : Y : Z = ⅙ : ⅛ : 1/24 = 4 : 3 : 1
Share of X = 600 * (4/8) = 300
Share of Y = 600 * (⅜) = 225
Share of Z = 600  (300+225) = 75
6. Another set of problem that you may face are the ones involving more than 2 different types of people.
For instance
2 women and 3 girls can do a piece of work in 10 days while 3 women and 2 girls can do the same work in 8 days. In how many days can 2 women and 1 girl do the work?
For such questions, let us assume that one woman’s 1day work = x and 1 girl’s 1day work = y
Now, 2x + 3y = 1/10 and 3x + 2y = ⅛
You now have two equations in two variables, solve the equations for the values of x and y and accordingly find the answer.
How to solve Time & Work (Shortcut Approach)
How to approach and express data while taking the exam?
If there are two people A and B. If A begins and both work alternate days. It means
1^{st} day  2^{nd} day  3^{rd} day  4^{th} day  5^{th}day  6^{th}day  7^{th}day  8^{th}day  9^{th}day  10^{th}day  ….. 
A  B  A  B  A  B  A  B  A  B  ….. 
Before going forward, we will discuss some basics of Ratios to make calculation easy for this topic and these basics will also help you in other topics like Partnership, time and distance, average, allegation etc.
TIME AND WORK EXAMPLES:
Example 5: A and B can do a work in 8 days and 12 days respectively.
A B
Time (T) 8 days 12 days
Efficiency (η) 12 8
η 3 2
total work = T_{A}× η_{A }= T_{B}× η_{B}total work = 8 × 3 = 12 × 2
total work = 24 units
(a). If both A and B work alternatively and A begins, then in how many days work will be completed?
According to the question, A begins and both and b work alternately.
Days  1^{st} day  2^{nd} day  3^{rd} day  4^{th} day  5^{th} day  6^{th} day  7^{th} day  8^{th} day  9^{th} day  10^{th }/2 day  Total Work 
Person  A  B  A  B  A  B  A  B  A  B 

Work Unit  3  2  3  2  3  2  3  2  3  1  24 
Hence work will be completed in 9 (1/2) days.
But this type of approach is not helpful in exams. We will go by basic but in a smarter way.
If we make a pair of A and B, we can say that both together will work 5 units but in 2 days.
2 days = 5 unit of work
2 days × 4 = 5 units of work × 4
8 Days = 20 units of work, i.e. we can say that up to 8^{th} day 20 units of work will be done
But on the 9th day it’s a chance of A and he will do his 3 units of work. So up to 9^{th} day 23 units of work will be done. Now 1 unit of work will remain. On 10^{th} day it’s a chance of B.
As above B do 2 unit of work in a day. So, he will do 1 unit of work in (1/2) days.
So total time taken to complete the work is 9 (1/2) days.
(b). If A and B both work for 4 days then A leaves. In how many days total work will be completed.
(Eff_{A+B}× t_{A+B })+( Eff._{B }× t_{B}) = total work
(5 × 4)+(2 × t_{B}) = 24
20 + 2 × t_{B }= 24
2 × t_{B} = 4
t_{B }= 2 days , hence total time will be ( 4+2 )days i.e. 6 days.
Important: the Same question can be framed in many ways.
Way 1: B works for 2 days and after that A joins B, then in how many days the work will be completed.
Solution: B works only for 2 days and then A joins B, i.e. both will work together after 2 days.
(t_{B }× η_{B}) + (t_{A+B}× η_{A+B}) = total work
(2 × 2) + (t_{A+B} × 5) = 24
(t_{A+B} × 5) = 20
t_{A+B } = 4 days hence total time is (2+4) days i.e. 6 days
Way 2: A and B both work together and A takes leaves for 2 days, then in how many days the work will be completed.
Solution: When both A and B are working together and A takes leaves for two days it means B has to work alone for 2 days .
Let total time to complete the work is t days.
So, η_{A+B }× (t2) + η_{B} × 2 days = 24
5 × (t2) + 2 × 2 = 24
t2 = 4
t = 6 days Hence total time taken to complete the work is 6 days.
Don’t confuse between t_{A+B }and T_{A+B }(as mentioned in article1). Both are different.
Example 6: X can do a work in 6 days, Y can do it in 8 days and Z can do it in 12 days.
(a). If X starts the work and X, Y, Z works in alternate days, then in how many days the work will be completed?
Solution: Here X will start work on the 1^{st} day, then Y will work on 2^{nd} day and z will work on the 3^{rd} day.
X : Y : Z
Time 6 : 8 : 12
Efficiency 12 × 8: 6 × 12 : 8 × 6
η 96 : 72 : 48
η 4 : 3 : 2
total work = η_{X}× T_{X}= η_{Y}×T_{Y }=η_{Z}×T_{Z} _{ }
total work = 6 × 4 = 8 × 3 = 12 × 2 = 24 units
On 1^{st} day, work done by X = 4 unit
On 2^{nd} day, work done by Y = 3 unit
On 3^{rd} day, work done by Z = 2 unit
Total work in 3 days done by X,Y and Z = 9 unit
Now again X will come then Y,then Z and so on till work is completed.
In 3 days = 9 units
× 2 × 2
In 6 days = 18 units
X will work on 7^{th} day = 4 units
= 22 units
Now we need (2422) units = 2 units work more but Y can do 3 unit of work in one day. So 2 unit will be done in (2/3) day.
+(2/3) day + 2 unit
Total days=7 (2/3) days 24 units
Hence total work will be done in 7 (2/3) days.
(b). If all started together and after completion of (3/4)^{th} work, Y left and remaining work is done by X and Z together. Then in how many days work will be completed?
Solution: Total work is 24 units then (3/4)^{th} work is 18 units.
One day work of (X+Y+Z) = 9 units so in 2 days 18 units of work will be done by (X+Y+Z) together.
After this Y left, X+Z worked together and 6 units of work remained.
One day work of X+Z = (4+2) units = 6 units. So in 3 days, total work will be completed.
How to solve the problems based on Men, Women and Children; Work and Wages
Man, Hours Efficiency Day Work Rupees Consumption etc. terms are used in these types of questions. We know that
No. of persons (M) ∝ Work (W)
efficiency (η) ∝ Work (W)
no. of days (D) ∝ Work (W)
no. of hours (H) ∝ Work (W)
So, No. of persons × efficiency × no. of days × no. of hours per day = work
M_{1}×η_{1}×D_{1}×H_{1} = W_{1} M_{2}×η_{2}×D_{2}×H_{2} = W_{2}
M_{1}×η_{1}×D_{1}×H_{1} = constant M_{2}×η_{2}×D_{2}×H_{2}_{ }= constant
W_{1} ………(1) W_{2} ………(2)
From eqn (1) and (2)
M_{1} × η_{1} × D_{1} × H_{1 }_{ }= M_{2} × η_{2} × D_{2} × H_{2}
W_{1} W_{2 }M_{1} × η_{1} × D_{1} × H_{1 }_{ }= M_{2} × η_{2} × D_{2} × H_{2}
W_{1}/R_{1}/C_{1 }W_{2}/R_{2}/C_{2}
where R= rupees or salary and C = consumption._{ }
Important: When persons are the same their efficiency will be equal. If persons are different their efficiency will be different. _{ }
Example 1: 12 persons can complete a work in 5 days, then how many persons are required to complete the same work in 3 days?
Solution: Here M_{1} = 12 and D_{1} = 5, D_{2 }= 3 days we have to find M_{2}
M_{1} × D_{1} = M_{2 }× D_{2 } _{ }12 × 5 = M_{2}× 3
60 = M_{2} × 3
M_{2} = 20
Example 2: 18 persons can make 12 chairs in 6 days working 8 hours per day. In many days 12 persons can make 24 chairs working 6 hours per day?
Solution: Here M_{1} = 18, D_{1} = 6, H_{1 }= 8 W_{1} = 12 and M_{2 }= 12,H_{2 }= 6, W_{2}=24 are given we have to find D_{2 }M_{1} × η_{1} × D_{1} × H_{1 }_{ }= M_{2} × η_{2} × D_{2} × H_{2}
W_{1 }W_{2 } 18 × 6 × 8 = 12 × D_{2 }× 6
12 24
D_{2} = 24 days
Example 3: The expenditure on fuel is Rs.600 burning 6 stoves for 12 days for 4 hours per day. How many stoves are required to burn 6 days for 8 hours making expenditure of Rs.900?
Solution: _{ Here R = Rupess}_{ }M_{1}×η_{1}×D_{1}×H_{1 }_{ }= M_{2}×η_{2}×D_{2}×H_{2}
R_{1} R_{2 }Here M = no. of stoves
6 × 12 × 4 = M_{2} × 6 × 8
600 900
M_{2} = 9 Hence, 9 stoves are required.
Example 4: If Q persons can do Q units of work in Q days working Q hours per day then in how many days P persons can do P units of work, working P hours per day.
Solution: M_{1} × D_{1} × H_{1 }_{ }= M_{2} × D_{2} × H_{2}
W_{1 }W_{2 } Q × Q × Q = P × D_{2 }× P
Q P_{ } D_{2} = Q^{2}_{ } days_{ } P
Example 5: If 5 women can do a work in 6 days working 9 hours per day. How many men are required to complete four times of work in 4 days working 6 hours per day? If 3 women can do the work in 6 hours that work can be done by 4 men in 3 hours.
Solution: In this type, we can see that the efficiencies of men and women are different. So first we will calculate efficiency, check the last line of question.
work was done by 3 women in 6 hours = work done by 4 men in 3 hours
3 ω × 6 = 4 m × 3
3ω = 2m or if ω = 2 then m = 3
ω = 2
m 3
where ω = eff. of women and m = effi. of men
using M_{1}×D_{1}×H_{1 }_{ }= M_{2}×D_{2}×H_{2}
W_{1 }W_{2}
5ω×6×9 = Xm×4×6
1 4_{ } 5×2×6×9 = X×3×4×6 (ω =2 and m = 3)_{ } 1 4
X = 30 men
Example 6: 4 women and 12 children together take 4 days to complete a piece of work. How many days will 4 children alone take to complete the piece of work if 2 women alone can complete the piece of work in 16 days?
Solution: Let time take by 4 children to complete the work is X days.
work done by 4 women and 12 children in 4 days =work done by 2 women in 16 days = work done by 4 children in X days
(4ω+12C) × 4 = (2ω) × 16 = (4C) × X
By using
(4ω+12C) × 4 = (2ω) × 16
16ω+48C = 32ω
48C = 16ω
3C = ω
hence ω = 3 and C = 1
By using
(2ω) × 16 = (4C) × X or (4ω+12C) × 4 = (4C) × X
2×3× 16 = 4× 1 × X or (4×3+12×1)× 4 = (4×1)× X
X = 24 days or (24) × 4 = 4 X
X = 24 days.
Important: There are so many shortcuts for this type of question, if the language of questions changes, students will be confused so we suggest you to go by this method. All approaches discussed above are the advanced part of Time and Work article 1 and 2.
Example 7: 4 men can complete a piece of work in 2 days. 4 women can complete the same piece of work in 4 days whereas 5 children can complete the same piece of work in 4 days. If 2 men, 4 women and 10 children work together, in how many days can the work be completed?
(SBI Rural Business officers)
Solution: Let time taken by (2m+4ω+10C) is X days.
4m × 2 = 4ω × 4 = 5C × 4 = (2m+4ω+10C) × X ……………..(1)
Firstly, we will calculate the efficiency of men, women and children.
8m = 16ω = 20C
2m = 4ω = 5C
m : ω : C (to know the basics of ratio, check time and work article 2)
20 : 10 : 8
10 : 5 : 4
Now, putting in eqn (1)
4×10×2 = 4×5×4 = 5×4×4 = (2×10+4×5+10×4)× X
80 = (80) X
X = 1 day.
Example 8: 8 men and 4 women together can complete a piece of work in 6 days. Work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days, 4 men left and 4 new women joined. In how many more days will the work be completed? (IBPS PO)
Solution: Given, work of man in 1 day = 2×work of a woman in 1 day
m × 1× 1 = 2 × ω × 1
m = 2ω So, m = 2 and ω = 1
Now, let total time take to complete the work is p days.
(8m+4ω)×6 = (8m+4ω)×2+(4m+8w)×(p2)
(16+4)× 6 = (16+4)× 2 +(8+8)× (p2)
120 = 40 + 16(p2)
80 = 16 (p2)
5 = p2
p = 7
hence total time is 7 days but in question time taken by changed persons after 2 days is asked so answer will be 5 days.
Exam approach: As (8m+4ω) has worked for 2 days after that 4m left and 4ω joined, remaining work of 4 days of (8m+4ω) will be done by (4m+8ω) in X days.
so, (8m+4ω) × 4 = (4m+8ω)× X
(16+4) × 4 = (8+8)X
80 = 16X
X = 5 days
Example 9: If 5 women or 3 men or 12 children can complete a work in 6 days. In how many days 2 men, 3 women and 5 children can complete the same work.?
Solution: In this question, we can see that here ‘OR’ is used instead of ‘AND’.
So, we can write directly 5ω = 3m = 12C
ω : m : C
12×3: 5×12: 5×3
12 : 20 : 5
5ω × 6 = 3m × 6 = 12C × 6 = (2m+3ω+5C)× X days
5× 12× 6 = 3×20×6 = 12×5×6 = (2×20+3×12+5×5)× X
360 = (40+36+25)X
X = 360/101 days.
Example 10: A contractor wants to complete a project in 120 days and he employed 80 men. After 90 days ½ work is completed, then how many more persons he must hire to complete work on time?
Solution: Here given work has to be completed in 120 days and 80 men were employed earlier.
According to contractor,
In 120 days = 1 total work completed
In 90 days = ¾ of work has to be completed but
we can see that it didn’t happen. In 90 days only ½ work is completed and the remaining ½ work has to be completed in the remaining 30 days.
Let more person hire are P.
90 × 80 = 30 × (P+80)
½ ½
240 = P+80
P = 160 persons
Example 11: P and Q undertake to do a piece of work in 6000Rs. P can do this in 8 days alone and B alone can do it in 12 days. With the help of R, they complete the work in 4 days. Find the part of P, Q and R individually?
Solution: Efficiency Days total work
24/8 = 3 P……………...8
24/12=2 Q…………….12 24
24/4 =6 P+Q+R......4 (LCM of 8,12and 4)
6  (3+2)= 1 R
there are two methods :
(1) efficiency method
The share of P = η_{P} × total Rs.
η_{(P+Q+R) } = 3 × 6000
6
= 3000 Rs.
The share of Q = 2 × 6000
6
= 2000 Rs.
The share of R = 1 × 6000
6
= 1000 Rs.
(2) Work Method
The share of P = work done by P × total Rs
work done by P+Q+R
= 3 × 4 × 6000
24
= 3000 Rs.
The share of Q = 2 × 4 × 6000
24
= 2000 Rs
Share of R = 1 × 4 × 6000
24
= 1000 Rs
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