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Work and Time Concepts
QUANTITATIVEAPTITUDE

Work and Time Concepts

Learn work-rate, efficiency, combined-work, and time-based techniques for aptitude problems.

1. Basic Work Rate

  • Formula:
1-day work=1Number of days \text{1-day work}=\frac1{\text{Number of days}}

If A completes a work in (x) days:

RA=1x R_A=\frac1x

Combined rate:

Rtotal=RA+RB+ R_{\text{total}}=R_A+R_B+\cdots
  • Example: A can complete a work in 12 days and B in 18 days. How long will they take together? Solution:
RA=112,RB=118 R_A=\frac1{12},\qquad R_B=\frac1{18} Rtotal=112+118=536 R_{\text{total}}=\frac1{12}+\frac1{18} =\frac5{36}

Therefore:

T=15/36=365=7.2 days T=\frac1{5/36} =\boxed{\frac{36}{5}=7.2\text{ days}}

2. Efficiency & Substitution

  • Formula:
Efficiency1Time \text{Efficiency}\propto\frac1{\text{Time}}

If efficiency ratio is (a:b), time ratio is:

b:a b:a
  • Example: A is 50% more efficient than B. B takes 30 days alone. How many days will they take together? Solution: A’s efficiency:
=150 =150%\text{ of B}=3:2

Therefore time ratio:

A:B=2:3 A:B=2:3

B takes 30 days, so A takes:

30×23=20 days 30\times\frac23=20\text{ days}

Together:

T=20×3020+30=12 days T=\frac{20\times30}{20+30} =\boxed{12\text{ days}}

3. Work Done in a Given Number of Days

  • Formula:
Work done=Rate×Time \text{Work done}=\text{Rate}\times\text{Time}

If A completes work in (x) days, work done in (d) days:

dx \frac d x
  • Example: A completes a work in 15 days. What fraction of the work does A complete in 6 days? Solution:
Work done=615=25 \text{Work done}=\frac6{15} =\boxed{\frac25}

4. Remaining Work After Partial Completion

  • Formula:
Remaining work=1Work completed \text{Remaining work}=1-\text{Work completed} Remaining time=Remaining workNew rate \text{Remaining time} =\frac{\text{Remaining work}}{\text{New rate}}
  • Example: A completes a work in 12 days. After working alone for 4 days, B joins and they finish the remaining work in 3 days. Find B’s individual time. Solution: A’s rate:
112 \frac1{12}

Work done by A in 4 days:

412=13 \frac4{12}=\frac13

Remaining:

113=23 1-\frac13=\frac23

A and B together complete (\frac23) in 3 days:

RA+B=2/33=29 R_{A+B}=\frac{2/3}{3}=\frac29

Therefore B’s rate:

29112=836336=536 \frac29-\frac1{12} =\frac8{36}-\frac3{36} =\frac5{36}

B’s time:

365=7.2 days \boxed{\frac{36}{5}=7.2\text{ days}}

5. Finding Individual Time from Combined Time

  • Formula: If A and B take (x) and (y) days:
1T=1x+1y \frac1T=\frac1x+\frac1y

Hence:

T=xyx+y T=\frac{xy}{x+y}
  • Example: A takes 8 days and A+B together take 4.8 days. Find B’s time. Solution:
1B=14.818 \frac1B=\frac1{4.8}-\frac18 =524324=224=112 =\frac5{24}-\frac3{24} =\frac2{24}=\frac1{12}

Therefore:

B=12 days \boxed{B=12\text{ days}}

6. Alternate Working Days

  • Formula:
Work in 2 days=1x+1y \text{Work in 2 days}=\frac1x+\frac1y

If A starts, calculate complete 2-day cycles first and then check the final day.

  • Example: A takes 10 days and B takes 15 days. They work on alternate days, starting with A. When is the work completed? Solution: Work in 2 days:
110+115=16 \frac1{10}+\frac1{15} =\frac16

After 10 days:

5×16=56 5\times\frac16=\frac56

Remaining:

16 \frac16

Day 11 is A’s turn:

110 \frac1{10}

Remaining:

16110=115 \frac16-\frac1{10} =\frac1{15}

Day 12 is B’s turn, and B completes:

115 \frac1{15}

Therefore work finishes on:

12th day \boxed{12\text{th day}}

7. Alternate Working with Unequal Final Day

  • Formula: Calculate work completed in complete cycles, then:
Final time=Complete cycles+Remaining workRate of next worker \text{Final time}= \text{Complete cycles}+\frac{\text{Remaining work}}{\text{Rate of next worker}}
  • Example: A completes a work in 6 days and B in 8 days. They work alternately starting with A. Find the completion time. Solution: Work in 2 days:
16+18=724 \frac16+\frac18=\frac7{24}

After 6 days (3 cycles):

3×724=78 3\times\frac7{24}=\frac78

Remaining:

18 \frac18

Day 7 is A’s turn. A’s rate:

16 \frac16

Time required:

1/81/6=34 day \frac{1/8}{1/6}=\frac34\text{ day}

Total:

634 days \boxed{6\frac34\text{ days}}

8. Men-Days / Workforce Problems

  • Formula: For the same work:
M1D1=M2D2 M_1D_1=M_2D_2

More generally:

WorkMen×Days \text{Work}\propto \text{Men}\times\text{Days}

when all workers have equal efficiency.

  • Example: 8 men complete a work in 15 days. How many days will 12 men take? Solution:
8×15=12×D 8\times15=12\times D D=12012=10 days D=\frac{120}{12} =\boxed{10\text{ days}}

9. Men-Women-Children Workforce Equations

  • Formula: Convert different workers into equivalent units using their efficiency ratio. If 1 woman does the work of (k) men:
Equivalent men=M+kW \text{Equivalent men}= M+kW

Then use:

Equivalent workers×Days=Constant \text{Equivalent workers}\times\text{Days}=\text{Constant}
  • Example: 2 women can do the work of 1 man. If 6 men complete a work in 10 days, how many days will 4 men and 4 women take? Solution: 4 women are equivalent to:
4×12=2 men 4\times\frac12=2\text{ men}

So total equivalent workers:

4+2=6 4+2=6

Since 6 men take 10 days:

10 days \boxed{10\text{ days}}

10. Workforce Change During a Project

  • Formula:
Total work=Workers×Days \text{Total work}=\text{Workers}\times\text{Days}

Calculate completed and remaining work separately when workers join or leave.

  • Example: 8 men complete a work in 15 days. After 5 days, 4 more men join. How many additional days are required? Solution: Total work:
8×15=120 man-days 8\times15=120\text{ man-days}

Work completed in 5 days:

8×5=40 8\times5=40

Remaining:

12040=80 120-40=80

New workforce:

8+4=12 8+4=12

Required days:

8012=623 days \frac{80}{12} =\boxed{6\frac23\text{ days}}

11. Workers Leaving During a Project

  • Formula:
Remaining work=New workforce×Remaining days \text{Remaining work} =\text{New workforce}\times\text{Remaining days}
  • Example: 12 men can complete a work in 20 days. After 8 days, 4 men leave. How many more days are required? Solution: Total work:
12×20=240 12\times20=240

Work completed:

12×8=96 12\times8=96

Remaining:

24096=144 240-96=144

Remaining workers:

124=8 12-4=8

Days required:

1448=18 days \frac{144}{8} =\boxed{18\text{ days}}

12. Wages Based on Work Done

  • Formula:
Individual wage======================Individual workTotal work×Total wage \text{Individual wage} ====================== \frac{\text{Individual work}}{\text{Total work}} \times\text{Total wage}

If workers have equal efficiency:

Wage ratio=Work ratio \text{Wage ratio}=\text{Work ratio}
  • Example: A, B and C work for 6, 4 and 5 days respectively at equal efficiency. If the total wage is ₹3,600, find A’s share. Solution: Work ratio:
6:4:5 6:4:5

Total:

15 15

A’s share:

3600×615=1440 3600\times\frac6{15} =\boxed{₹1440}

13. Wages Based on Efficiency × Time

  • Formula:
Wage ratio=E1T1:E2T2 \text{Wage ratio} =E_1T_1:E_2T_2
  • Example: A works 6 days at efficiency 2 units/day, while B works 4 days at efficiency 3 units/day. If ₹2,400 is distributed according to work, find their shares. Solution: A’s work:
2×6=12 2\times6=12

B’s work:

3×4=12 3\times4=12

Ratio:

1:1 1:1

Each gets:

1200 \boxed{₹1200}

14. Efficiency Ratio and Time Ratio

  • Formula:
EA:EB=TB:TA E_A:E_B=T_B:T_A

If A is (x%) more efficient than B:

EA:EB=(100+x):100 E_A:E_B=(100+x):100
  • Example: A is 25% more efficient than B. If B takes 20 days, how many days does A take? Solution:
EA:EB=125:100=5:4 E_A:E_B=125:100=5:4

Therefore:

TA:TB=4:5 T_A:T_B=4:5 TA=20×45=16 days T_A=20\times\frac45 =\boxed{16\text{ days}}

15. Work Equivalence from Efficiency

  • Formula:
Work doneEfficiency×Time \text{Work done}\propto\text{Efficiency}\times\text{Time}

Therefore:

E1T1=E2T2 E_1T_1=E_2T_2

for the same amount of work.

  • Example: A is twice as efficient as B. If A works for 6 days, how many days must B work to complete the same amount of work? Solution:
EA=2EB E_A=2E_B

Thus:

2EB×6=EB×T 2E_B\times6=E_B\times T T=12 T=12

Therefore:

12 days \boxed{12\text{ days}}

Advanced Variants

16. A, B and C Working Together

  • Formula:
RA+B+C=RA+RB+RC R_{A+B+C}=R_A+R_B+R_C

Hence:

T=1RA+RB+RC T=\frac1{R_A+R_B+R_C}
  • Example: A, B and C can complete a work in 12, 18 and 36 days respectively. How long will they take together? Solution:
R=112+118+136 R=\frac1{12}+\frac1{18}+\frac1{36} =336+236+136=636=16 =\frac3{36}+\frac2{36}+\frac1{36} =\frac6{36}=\frac16

Therefore:

6 days \boxed{6\text{ days}}

17. One Worker Joins or Leaves at a Specific Time

  • Formula:
Total work=Work in first phase+Work in second phase \text{Total work} =\text{Work in first phase}+\text{Work in second phase}

For each phase:

W=R×T W=R\times T
  • Example: A completes a work in 12 days and B in 18 days. A works alone for 3 days, then B joins. Find the total completion time. Solution: A’s rate:
112 \frac1{12}

Work in 3 days:

312=14 \frac3{12}=\frac14

Remaining:

34 \frac34

Combined rate:

112+118=536 \frac1{12}+\frac1{18}=\frac5{36}

Time for remaining work:

3/45/36=34×365=275=5.4 \frac{3/4}{5/36} =\frac34\times\frac{36}{5} =\frac{27}{5}=5.4

Total:

3+5.4=8.4 days 3+5.4=\boxed{8.4\text{ days}}

18. Pipes and Workers Combined

  • Formula: Treat every entity as a work rate:
Rnet=Rworkers+RfillingRemptying R_{\text{net}}=R_{\text{workers}}+R_{\text{filling}}-R_{\text{emptying}}
  • Example: A worker completes a job in 10 days. A machine completes it in 15 days. A faulty process removes (\frac1{30}) of the work per day. How long do they take together? Solution:
R=110+115130 R=\frac1{10}+\frac1{15}-\frac1{30} =330+230130=430=215 =\frac3{30}+\frac2{30}-\frac1{30} =\frac4{30}=\frac2{15}

Therefore:

T=152=7.5 days T=\frac{15}{2} =\boxed{7.5\text{ days}}

19. Work Completed by Alternating Groups

  • Formula: Calculate the work of each cycle:
Wcycle=R1t1+R2t2 W_{\text{cycle}}=R_1t_1+R_2t_2

Then determine complete cycles and the remaining work.

  • Example: A can finish a work in 8 days and B in 12 days. A works for 2 days, B for 1 day repeatedly. How long will the work take? Solution: Work in one 3-day cycle:
2(18)+1(112)=14+112=13 2\left(\frac18\right)+1\left(\frac1{12}\right) =\frac14+\frac1{12} =\frac13

After 2 cycles (6 days):

23 \frac23

Remaining:

13 \frac13

A needs:

1/31/8=83 \frac{1/3}{1/8}=\frac83

days.

Total:

6+83=823 days6+\frac83 =\boxed{8\frac23\text{ days}}

20. Work and Wages with Different Efficiencies

  • Formula:
Wage ratio=E1T1:E2T2 \text{Wage ratio} =E_1T_1:E_2T_2

Do not use only the number of days when efficiencies differ.

  • Example: A works for 5 days at efficiency 3 units/day and B works for 6 days at efficiency 2 units/day. Total wage is ₹2,100. Find A’s share. Solution: A’s work:
5×3=15 5\times3=15

B’s work:

6×2=12 6\times2=12

Ratio:

15:12=5:4 15:12=5:4

A’s share:

2100×59=1166.67 2100\times\frac59 =\boxed{₹1166.67}

B’s share:

933.33 \boxed{₹933.33}

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