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Quantitative Aptitude · Time and Work

Basics- Concepts, Formulas and Key Points

Explanation

🟠 1. BASIC CONCEPT Time and Work problems are based on the relationship between:
Work,Time,Efficiency \text{Work},\quad \text{Time},\quad \text{Efficiency}
If a person completes a work in TT days, then the work done by that person in 1 day is:
One Day’s Work=1T \text{One Day's Work}=\frac{1}{T}
If a person completes a work in 10 days:
One Day’s Work=110 \text{One Day's Work}=\frac{1}{10}
🟠 2. BASIC WORK FORMULA
Work=Rate×Time \text{Work}=\text{Rate}\times\text{Time}
Therefore:
Rate=WorkTime \text{Rate}=\frac{\text{Work}}{\text{Time}}
Time=WorkRate \text{Time}=\frac{\text{Work}}{\text{Rate}}
🟠 3. ONE DAY'S WORK If A can complete a work in xx days:
A′s one day work=1x A's\ one\ day\ work=\frac{1}{x}
If B can complete the same work in yy days:
B′s one day work=1y B's\ one\ day\ work=\frac{1}{y}
🟠 4. WORK DONE TOGETHER If A completes a work in xx days and B completes it in yy days, their combined one-day work is:
1x+1y \frac{1}{x}+\frac{1}{y}
Therefore, time taken together is:
T=11x+1y T=\frac{1}{\frac{1}{x}+\frac{1}{y}}
Simplifying:
T=xyx+y T=\frac{xy}{x+y}
🟠 5. THREE PEOPLE WORKING TOGETHER If A, B and C complete the work individually in xx, yy and zz days:
Combined One Day’s Work=1x+1y+1z \text{Combined One Day's Work} = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
Therefore:
T=11x+1y+1z T= \frac{1} {\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}
🟠 6. EFFICIENCY Efficiency means the amount of work completed per unit of time.
Efficiency=WorkTime \text{Efficiency} = \frac{\text{Work}}{\text{Time}}
For the same amount of work:
Efficiency∝1Time \text{Efficiency}\propto\frac{1}{\text{Time}}
🟠 7. EFFICIENCY AND TIME RELATION If A and B have efficiencies in the ratio:
A:B=m:n A:B=m:n
Then their times taken to complete the same work are in the inverse ratio:
A′s Time:B′s Time=n:m A's\ Time:B's\ Time=n:m
🟠 8. WORK RATIO If A and B work for the same amount of time and their efficiencies are in the ratio:
A:B=m:n A:B=m:n
Then the work done is also in the ratio:
m:n m:n
🟠 9. WORK DONE IN GIVEN DAYS If a person can complete a work in TT days, then work completed in DD days is:
DT \frac{D}{T}
of the total work. 🟠 10. REMAINING WORK If a person has completed xy\frac{x}{y} of a work, then the remaining work is:
1−xy 1-\frac{x}{y}
=y−xy =\frac{y-x}{y}
🟠 11. WORK AND WAGES If workers have different efficiencies, wages should be divided according to the work done.
Wages∝Work Done \text{Wages}\propto\text{Work Done}
If the workers work for the same time:
Wages∝Efficiency \text{Wages}\propto\text{Efficiency}
🟠 12. MEN AND DAYS If the amount of work is fixed:
Men×Days=Constant \text{Men}\times\text{Days}=\text{Constant}
Therefore:
M1D1=M2D2 M_1D_1=M_2D_2
More men require fewer days. Fewer men require more days. 🟠 13. MEN, DAYS AND HOURS If the number of working hours per day also changes:
M1D1H1=M2D2H2 M_1D_1H_1=M_2D_2H_2
where:
M=Number of men M=\text{Number of men}
D=Number of days D=\text{Number of days}
H=Working hours per day H=\text{Working hours per day}
🟠 14. MEN, DAYS AND EFFICIENCY If efficiency also changes:
M1D1E1=M2D2E2 M_1D_1E_1=M_2D_2E_2
This is useful when workers have different efficiencies. 🟠 15. WORK AND TIME For the same amount of work:
Time∝1Efficiency \text{Time}\propto\frac{1}{\text{Efficiency}}
If efficiency increases, time decreases. If efficiency decreases, time increases. 🟠 16. A WORKS ALONE AND B JOINS LATER If A completes xy\frac{x}{y} of the work before B joins, then remaining work is:
1−xy 1-\frac{x}{y}
Then calculate the combined rate of A and B for the remaining work. 🟠 17. A LEAVES BEFORE COMPLETION If A and B work together initially and A leaves after a certain number of days:
Work Done=Combined Rate×Time \text{Work Done} = \text{Combined Rate}\times\text{Time}
Then:
Remaining Work=Total Work−Work Done \text{Remaining Work} = \text{Total Work}-\text{Work Done}
The remaining work is completed by the remaining worker. 🟠 18. ALTERNATE WORKING If A and B work on alternate days:
Work in one cycle=A’s one-day work+B’s one-day work \text{Work in one cycle} = \text{A's one-day work}+\text{B's one-day work}
For two-day cycles:
Cycle Work=1x+1y \text{Cycle Work} = \frac{1}{x}+\frac{1}{y}
🟠 19. PIPES AND CISTERNS The same Time and Work concept applies to pipes. If a pipe fills a tank in xx hours:
Filling Rate=1x \text{Filling Rate}=\frac{1}{x}
If another pipe empties the tank in yy hours:
Emptying Rate=−1y \text{Emptying Rate}=-\frac{1}{y}
Net rate:
1x−1y \frac{1}{x}-\frac{1}{y}
🟠 20. FILLING PIPES TOGETHER If two pipes fill a tank in xx and yy hours:
Combined Rate=1x+1y \text{Combined Rate} = \frac{1}{x}+\frac{1}{y}
Therefore:
T=xyx+y T=\frac{xy}{x+y}
🟠 21. FILLING AND EMPTYING PIPE TOGETHER If a pipe fills a tank in xx hours and another empties it in yy hours:
Net Rate=1x−1y \text{Net Rate} = \frac{1}{x}-\frac{1}{y}
Therefore:
T=11x−1y T= \frac{1} {\frac{1}{x}-\frac{1}{y}}
🟠 22. LCM METHOD A convenient method for solving Time and Work problems is to assume the total work as the LCM of the individual times. For example, if A takes 10 days and B takes 15 days:
LCM(10,15)=30 LCM(10,15)=30
Assume:
Total Work=30 units \text{Total Work}=30\text{ units}
A's efficiency:
3010=3 units/day \frac{30}{10}=3\text{ units/day}
B's efficiency:
3015=2 units/day \frac{30}{15}=2\text{ units/day}
🟠 23. IMPORTANT KEY POINTS • Always convert the given information into work done per unit time. • If a person takes more days, their efficiency is lower. • If a person takes fewer days, their efficiency is higher. • For people working together, add their individual work rates. • For a person leaving the work, calculate the work completed before leaving and then calculate the remaining work. • For fixed work:
Men×Days=Constant \text{Men}\times\text{Days}=\text{Constant}
• When hours per day are included:
M×D×H=Constant M\times D\times H=\text{Constant}
• When efficiency is included:
M×D×E=Constant M\times D\times E=\text{Constant}
• Pipes filling a tank are treated as positive work. • Pipes emptying a tank are treated as negative work. • In alternate-day problems, calculate the work completed in one complete cycle. • LCM method often makes calculations easier.

Example

🟢 EXAMPLE 1: ONE DAY'S WORK QUESTION: A can complete a work in 10 days. What part of the work does A complete in one day? SOLUTION:
One Day’s Work=110 \text{One Day's Work}=\frac{1}{10}
ANSWER:
110 \frac{1}{10}
🟢 EXAMPLE 2: TWO PEOPLE WORKING TOGETHER QUESTION: A can complete a work in 10 days and B can complete it in 15 days. How many days will they take together? SOLUTION: A's one-day work:
110 \frac{1}{10}
B's one-day work:
115 \frac{1}{15}
Combined work:
110+115 \frac{1}{10}+\frac{1}{15}
Taking LCM:
=3+230 =\frac{3+2}{30}
=530 =\frac{5}{30}
=16 =\frac{1}{6}
Therefore:
T=6 days T=6\text{ days}
ANSWER:
6 days 6\text{ days}
🟢 EXAMPLE 3: THREE PEOPLE WORKING TOGETHER QUESTION: A can complete a work in 12 days, B in 18 days and C in 36 days. Find the time taken by all three together. SOLUTION: Combined one-day work:
112+118+136 \frac{1}{12}+\frac{1}{18}+\frac{1}{36}
Taking LCM 36:
=3+2+136 =\frac{3+2+1}{36}
=636 =\frac{6}{36}
=16 =\frac{1}{6}
Therefore:
T=6 days T=6\text{ days}
ANSWER:
6 days 6\text{ days}
🟢 EXAMPLE 4: EFFICIENCY RATIO QUESTION: A is twice as efficient as B. If B completes a work in 20 days, how many days will A take? SOLUTION: Since A is twice as efficient:
A:B=2:1 A:B=2:1
Time is inversely proportional to efficiency. Therefore:
A′s Time:B′s Time=1:2 A's\ Time:B's\ Time=1:2
Hence:
A′s Time=202 A's\ Time=\frac{20}{2}
=10 days =10\text{ days}
ANSWER:
10 days 10\text{ days}
🟢 EXAMPLE 5: FINDING WORK DONE QUESTION: A can complete a work in 20 days. What fraction of the work will A complete in 8 days? SOLUTION: One-day work:
120 \frac{1}{20}
Work done in 8 days:
820 \frac{8}{20}
=25 =\frac{2}{5}
ANSWER:
25 \frac{2}{5}
🟢 EXAMPLE 6: REMAINING WORK QUESTION: A completes 35\frac{3}{5} of a work. What fraction of the work remains? SOLUTION:
Remaining Work=1−35 \text{Remaining Work} = 1-\frac{3}{5}
=5−35 =\frac{5-3}{5}
=25 =\frac{2}{5}
ANSWER:
25 \frac{2}{5}
🟢 EXAMPLE 7: A WORKS ALONE, THEN B JOINS QUESTION: A can complete a work in 12 days. A works alone for 4 days, after which B joins A. If B alone can complete the work in 24 days, how many more days are required to finish the work? SOLUTION: A's one-day work:
112 \frac{1}{12}
Work done by A in 4 days:
412 \frac{4}{12}
=13 =\frac{1}{3}
Remaining work:
1−13 1-\frac{1}{3}
=23 =\frac{2}{3}
B's one-day work:
124 \frac{1}{24}
Combined one-day work:
112+124 \frac{1}{12}+\frac{1}{24}
=324 =\frac{3}{24}
=18 =\frac{1}{8}
Time required:
T=2/31/8 T=\frac{2/3}{1/8}
=23×8 =\frac{2}{3}\times8
=163 =\frac{16}{3}
=513 days =5\frac{1}{3}\text{ days}
ANSWER:
513 days 5\frac{1}{3}\text{ days}
🟢 EXAMPLE 8: A LEAVES AFTER WORKING TOGETHER QUESTION: A and B can complete a work in 12 and 18 days respectively. They work together for 3 days, after which A leaves. How many more days will B take to finish the remaining work? SOLUTION: Combined one-day work:
112+118 \frac{1}{12}+\frac{1}{18}
=536 =\frac{5}{36}
Work completed in 3 days:
3×536 3\times\frac{5}{36}
=512 =\frac{5}{12}
Remaining work:
1−512 1-\frac{5}{12}
=712 =\frac{7}{12}
B's one-day work:
118 \frac{1}{18}
Time required:
T=7/121/18 T=\frac{7/12}{1/18}
=712×18 =\frac{7}{12}\times18
=212 =\frac{21}{2}
=1012 days =10\frac{1}{2}\text{ days}
ANSWER:
1012 days 10\frac{1}{2}\text{ days}
🟢 EXAMPLE 9: MEN AND DAYS QUESTION: 12 men can complete a work in 15 days. How many days will 20 men take to complete the same work? SOLUTION: For fixed work:
M1D1=M2D2 M_1D_1=M_2D_2
12×15=20×D2 12\times15=20\times D_2
180=20D2 180=20D_2
D2=9 D_2=9
ANSWER:
9 days 9\text{ days}
🟢 EXAMPLE 10: MEN AND HOURS QUESTION: 10 men working 8 hours per day can complete a work in 12 days. How many days will 16 men working 6 hours per day take? SOLUTION: Using:
M1D1H1=M2D2H2 M_1D_1H_1=M_2D_2H_2
10×12×8=16×D2×6 10\times12\times8 = 16\times D_2\times6
960=96D2 960=96D_2
D2=10 D_2=10
ANSWER:
10 days 10\text{ days}
🟢 EXAMPLE 11: EFFICIENCY AND MEN QUESTION: 8 men can complete a work in 15 days. If each new worker is 25% more efficient than an original worker, how many days will 10 such workers take? SOLUTION: Efficiency of each new worker:
100%+25%=125% 100\%+25\%=125\%
Thus:
10×125=1250 10\times125 = 1250
Equivalent efficiency of 10 new workers compared with original workers:
1250100=12.5 \frac{1250}{100}=12.5
Using:
M1D1=M2D2 M_1D_1=M_2D_2
where effective workers are 8 and 12.5:
8×15=12.5×D2 8\times15=12.5\times D_2
120=12.5D2 120=12.5D_2
D2=9.6 D_2=9.6
ANSWER:
9.6 days 9.6\text{ days}
🟢 EXAMPLE 12: WAGES QUESTION: A and B complete a work together and receive ₹1,200. A does 23\frac{2}{3} of the work and B does 13\frac{1}{3}. Find their respective shares. SOLUTION: Wages are proportional to work done. A's share:
1200×23 1200\times\frac{2}{3}
=800 =800
B's share:
1200×13 1200\times\frac{1}{3}
=400 =400
ANSWER:
A=₹800 A=₹800
B=₹400 B=₹400
🟢 EXAMPLE 13: ALTERNATE DAYS QUESTION: A can complete a work in 10 days and B can complete it in 15 days. They work on alternate days, starting with A. How many days will they take to complete the work? SOLUTION: A's one-day work:
110 \frac{1}{10}
B's one-day work:
115 \frac{1}{15}
Work completed in 2 days:
110+115 \frac{1}{10}+\frac{1}{15}
=16 =\frac{1}{6}
In 10 days, there are 5 complete cycles. Work completed:
5×16 5\times\frac{1}{6}
=56 =\frac{5}{6}
Remaining work:
1−56 1-\frac{5}{6}
=16 =\frac{1}{6}
A works on the 11th day:
110 \frac{1}{10}
Remaining after A:
16−110 \frac{1}{6}-\frac{1}{10}
=115 =\frac{1}{15}
B completes this remaining work in one day. Therefore:
T=12 days T=12\text{ days}
ANSWER:
12 days 12\text{ days}
🟢 EXAMPLE 14: PIPES FILLING TOGETHER QUESTION: A pipe can fill a tank in 12 hours and another pipe can fill it in 18 hours. How long will they take together? SOLUTION: First pipe's rate:
112 \frac{1}{12}
Second pipe's rate:
118 \frac{1}{18}
Combined rate:
112+118 \frac{1}{12}+\frac{1}{18}
=536 =\frac{5}{36}
Therefore:
T=365 T=\frac{36}{5}
=7.2 hours =7.2\text{ hours}
ANSWER:
7.2 hours 7.2\text{ hours}
🟢 EXAMPLE 15: FILLING AND EMPTYING PIPE QUESTION: A pipe fills a tank in 10 hours while another pipe empties it in 15 hours. If both are opened together, how long will it take to fill the tank? SOLUTION: Filling rate:
110 \frac{1}{10}
Emptying rate:
115 \frac{1}{15}
Net rate:
110−115 \frac{1}{10}-\frac{1}{15}
=3−230 =\frac{3-2}{30}
=130 =\frac{1}{30}
Therefore:
T=30 hours T=30\text{ hours}
ANSWER:
30 hours 30\text{ hours}
🟢 EXAMPLE 16: TWO WORKERS WITH DIFFERENT EFFICIENCIES QUESTION: A is 50% more efficient than B. If B can complete a work in 18 days, how many days will A take? SOLUTION: Let B's efficiency be:
100 100
A's efficiency:
100+50=150 100+50=150
Efficiency ratio:
A:B=150:100 A:B=150:100
=3:2 =3:2
Time ratio is inverse:
A:B=2:3 A:B=2:3
Therefore:
A′s Time=18×23 A's\ Time = 18\times\frac{2}{3}
=12 days =12\text{ days}
ANSWER:
12 days 12\text{ days}
🟢 EXAMPLE 17: FINDING ONE WORKER'S TIME QUESTION: A and B together can complete a work in 8 days. A alone can complete it in 12 days. How many days will B alone take? SOLUTION: Combined one-day work:
18 \frac{1}{8}
A's one-day work:
112 \frac{1}{12}
B's one-day work:
18−112 \frac{1}{8}-\frac{1}{12}
=3−224 =\frac{3-2}{24}
=124 =\frac{1}{24}
Therefore:
B′s Time=24 days B's\ Time=24\text{ days}
ANSWER:
24 days 24\text{ days}
🟢 EXAMPLE 18: PART OF WORK COMPLETED QUESTION: A can complete a work in 16 days. B can complete the same work in 24 days. They work together for 4 days. What fraction of the work remains? SOLUTION: Combined one-day work:
116+124 \frac{1}{16}+\frac{1}{24}
=3+248 =\frac{3+2}{48}
=548 =\frac{5}{48}
Work completed in 4 days:
4×548 4\times\frac{5}{48}
=512 =\frac{5}{12}
Remaining work:
1−512 1-\frac{5}{12}
=712 =\frac{7}{12}
ANSWER:
712 \frac{7}{12}
🟢 EXAMPLE 19: WORK AND EFFICIENCY QUESTION: A is 25% more efficient than B. If B takes 20 days to complete a work, find the time taken by A. SOLUTION: B's efficiency:
100% 100\%
A's efficiency:
125% 125\%
Therefore:
A:B=125:100 A:B=125:100
=5:4 =5:4
Time ratio:
A:B=4:5 A:B=4:5
Therefore:
A′s Time=20×45 A's\ Time=20\times\frac{4}{5}
=16 days =16\text{ days}
ANSWER:
16 days 16\text{ days}
🟢 EXAMPLE 20: MEN, DAYS AND HOURS QUESTION: 15 men working 6 hours per day complete a work in 20 days. How many men are required to complete the same work in 15 days by working 8 hours per day? SOLUTION: Using:
M1D1H1=M2D2H2 M_1D_1H_1=M_2D_2H_2
15×20×6=M2×15×8 15\times20\times6 = M_2\times15\times8
1800=120M2 1800=120M_2
M2=15 M_2=15
ANSWER:
15 men 15\text{ men}