Time and Work

Time and Work: Complete Beginner Guide with Detailed Examples

Time and Work studies how fast a person, worker, machine or pipe completes a job. The work is the whole job, the time tells us how long it takes, and the rate or efficiency tells us how much of the job is completed in one unit of time.

Suppose a worker finishes one wall in 10 days. If the worker does the same amount each day, then one day completes one of the 10 equal parts of the wall. The worker's one-day work is therefore one-tenth of the wall. Five days complete five-tenths, which is half of the wall.

This one-day-work idea connects the entire topic. We will begin with it, then learn how rates combine, why efficiency and time move in opposite directions, how unfinished work is handled, and how the same model solves workers, machines, wages, pipes and changing schedules.

The Three Basic Quantities

The topic is built from three connected ideas:

Quantity	Easy meaning
Work	The full task or the part of it being discussed
Rate or efficiency	Work completed in one unit of time
Time	Number of time units spent working

Their basic relationship is:

$$\text{Work}=\text{Rate}\times \text{Time}$$

This formula is not special to workers. Distance equals speed multiplied by time for the same reason. Production equals production rate multiplied by time. A tank's filled part equals filling rate multiplied by time.

From the same relationship:

$$\text{Rate}=\frac{\text{Work}}{\text{Time}}$$ $$\text{Time}=\frac{\text{Work}}{\text{Rate}}$$

We divide work by rate when finding time because we are asking how many one-time-unit pieces are needed to make the required work.

Why We Usually Treat the Whole Work as 1

A job does not always have to be counted in bricks, pages, units or tasks. When only the complete job matters, we can safely call that whole job 1.

The number 1 does not mean one brick or one hour. It means one complete job, which is 100 percent of whatever task is being discussed.

If a person finishes the whole job in $D$ days, the one-day work is:

$$\text{One-day work}=\frac{1}{D}$$

This works because one full job is shared equally across $D$ days. For 18 days, each day contributes one of 18 equal parts.

Case 1: Find the work completed in some days

Example: A worker can complete a job in 18 days. What fraction of the job will be completed in 5 days?

What is given? One full job takes 18 days. The worker keeps the same rate each day.

What are we finding? We need the fraction completed after only 5 days, not the time for the whole work.

First find the work in one day:

$$\text{One-day work}=\frac{1}{18}$$

Five days contain five equal daily work parts, so multiply by 5:

$$\text{Work in 5 days}=5\times \frac{1}{18}=\frac{5}{18}$$

The worker completes $\frac{5}{18}$ of the job. The remaining part would be $1-\frac{5}{18}=\frac{13}{18}$.

Case 2: Find the full time from partial work

Example: A completes $\frac{5}{8}$ of a job in 10 days. How long will the whole job take at the same rate?

The 10 days belong to only five-eighths of the work. First find the work completed in one day by sharing that fraction equally across 10 days:

$$\text{One-day work}=\frac{\frac{5}{8}}{10}=\frac{5}{8}\times \frac{1}{10}=\frac{1}{16}$$

A completes one-sixteenth of the whole job per day. A full job contains 16 such parts, so:

$$\text{Full time}=\frac{1}{\frac{1}{16}}=16\text{days}$$

The whole job takes 16 days. We divided by the completed fraction because 10 days represents only part of the full time.

Case 3: Another way to scale partial work

Example: A completes $\frac{3}{7}$ of a job in 9 days. Find the full time.

Think of the job as 7 equal parts. Three parts take 9 days, so one part takes:

$$\frac{9}{3}=3\text{days}$$

The full job has 7 such parts:

$$7\times 3=21\text{days}$$

The full job takes 21 days. This is the same as $\frac{9}{\frac{3}{7}}$.

Adding Rates When People Work Together

When two people help on the same job at the same time, their completed parts are added. If A completes one-fifteenth and B completes one-twentieth in a day, together they complete both parts in that day.

For workers with individual times $a$, $b$ and $c$:

$$\text{Combined rate}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

Once the combined one-day work is known, time is full work divided by that rate:

$$\text{Combined time}=\frac{1}{\text{Combined rate}}$$

Case 1: Two people working together

Example: A finishes a task in 15 days and B finishes it in 20 days. How long will they take together?

A's one-day work is $\frac{1}{15}$, and B's is $\frac{1}{20}$. They help each other, so add:

$$\frac{1}{15}+\frac{1}{20}$$

The least common denominator is 60. This lets us describe both rates with equal-sized pieces:

$$\frac{1}{15}=\frac{4}{60},\frac{1}{20}=\frac{3}{60}$$

Together:

$$\text{Combined rate}=\frac{4}{60}+\frac{3}{60}=\frac{7}{60}$$

They complete 7 of 60 equal work parts each day. To complete all 60 parts:

$$\text{Time}=\frac{1}{\frac{7}{60}}=\frac{60}{7}=8\frac{4}{7}\text{days}$$

Together they take $8\frac{4}{7}$ days. This is less than both 15 and 20 days, which is a useful reason check.

Case 2: Three people working together

Example: A, B and C can complete a job alone in 12, 18 and 36 days. Find their combined time.

Their one-day rates are $\frac{1}{12}$, $\frac{1}{18}$ and $\frac{1}{36}$. Use 36 as a common denominator:

$$\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{6}{36}=\frac{1}{6}$$

Together they complete one-sixth of the job each day. Six equal days complete six-sixths:

$$\text{Time}=\frac{1}{\frac{1}{6}}=6\text{days}$$

The three workers take 6 days.

The LCM Work-Unit Method

Fractions are correct, but some learners find whole work units easier. We may choose the total work as the least common multiple of the individual times. This changes only the measuring unit, not the real job.

For A taking 12 days and B taking 18 days, choose total work as $operatornameLCM\left(\mathrm{12,18}\right)=36$ units. Then:

• A completes $\frac{36}{12}=3$ units per day
• B completes $\frac{36}{18}=2$ units per day
• together they complete $3+2=5$ units per day

Their time is:

$$\frac{36\text{units}}{5\text{units per day}}=\frac{36}{5}=7\frac{1}{5}\text{days}$$

This is exactly the same as adding $\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$. The fraction method calls the work 1. The LCM method calls it 36 units. Both describe the same rate.

Finding a Missing Worker's Rate

Sometimes we know the completion time of a full team and the individual times of only some members. The full team's rate contains everyone's work. To isolate the missing worker, subtract the known rates.

Case 1: Find C's individual time

Example: A takes 12 days, B takes 18 days, and A, B and C together take 5 days. Find C's time alone.

The full team's one-day work is:

$$\frac{1}{5}$$

A and B together complete:

$$\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$$

The full-team rate equals A's rate plus B's rate plus C's rate. Therefore, remove the known A and B part:

$$C\text{'s rate}=\frac{1}{5}-\frac{5}{36}=\frac{36-25}{180}=\frac{11}{180}$$

C completes eleven of 180 equal parts each day. The time for one complete job is the reciprocal:

$$C\text{'s time}=\frac{1}{\frac{11}{180}}=\frac{180}{11}=16\frac{4}{11}\text{days}$$

C alone takes $16\frac{4}{11}$ days.

Case 2: A known individual and a known pair

Example: A alone takes 30 days, while A and B together take 20 days. Find B's individual time.

A's rate is $\frac{1}{30}$ and the pair's rate is $\frac{1}{20}$. B's contribution is what remains after A's contribution is removed:

$$B\text{'s rate}=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}=\frac{1}{60}$$

B completes one-sixtieth of the job per day, so B alone takes 60 days.

It may feel unusual that B is slower than A even though the pair takes only 20 days. The numbers are consistent: A adds one-thirtieth and B adds one-sixtieth, giving three-sixtieths, which equals one-twentieth.

Work Completed and Work Remaining

When people work for only part of the schedule, first calculate what they finish. Then subtract from the full work:

$$\text{Remaining work}=1-\text{Completed work}$$

Case 1: Find the unfinished fraction

Example: A and B can finish a job alone in 18 and 24 days. They work together for 4 days. What fraction remains?

Their combined one-day work is:

$$\frac{1}{18}+\frac{1}{24}=\frac{4}{72}+\frac{3}{72}=\frac{7}{72}$$

Four days complete four copies of that daily work:

$$\text{Completed work}=4\times \frac{7}{72}=\frac{28}{72}=\frac{7}{18}$$

The full work is 1, so:

$$\text{Remaining work}=1-\frac{7}{18}=\frac{11}{18}$$

The unfinished part is $\frac{11}{18}$.

Case 2: One person starts and another finishes

Example: A takes 30 days and works for 12 days. B, who can finish the full job in 40 days, completes the rest. Find the total time.

A completes this fraction in 12 days:

$$12\times \frac{1}{30}=\frac{12}{30}=\frac{2}{5}$$

Therefore, the remaining work is:

$$1-\frac{2}{5}=\frac{3}{5}$$

B's rate is $\frac{1}{40}$. Time for B to complete only three-fifths is:

$$\frac{\frac{3}{5}}{\frac{1}{40}}=\frac{3}{5}\times 40=24\text{days}$$

A worked for 12 days and B worked for 24 more days:

$$\text{Total time}=12+24=36\text{days}$$

The job finishes 36 days after A began.

Case 3: One worker leaves after working together

Example: A takes 20 days and B takes 30 days. They work together for 8 days, then A leaves. How many more days does B need?

Together, their rate is:

$$\frac{1}{20}+\frac{1}{30}=\frac{1}{12}$$

In 8 days, they complete:

$$8\times \frac{1}{12}=\frac{2}{3}$$

One-third remains. B completes one-thirtieth per day, so:

$$\text{Extra time}=\frac{\frac{1}{3}}{\frac{1}{30}}=10\text{days}$$

B needs 10 more days after A leaves. If we need the complete time from the start, it is $8+10=18$ days.

Efficiency and Time Are Inversely Related

Efficiency means work completed per day. A more efficient worker completes a larger part each day, so fewer days are needed for the same work.

For the same job:

$$\text{Efficiency}\times \text{Time}=\text{constant}$$

Therefore:

$$E_1:E_2=T_2:T_1$$

The time ratio is reversed because efficiency and time move in opposite directions.

Case 1: One worker is twice as efficient

Example: A is twice as efficient as B and finishes 30 days earlier. Find their combined time.

If A's efficiency is twice B's, A takes half as much time. Let A take $x$ days. Then B takes $2x$ days.

Their time difference is 30 days:

$$2x-x=30$$ $$x=30$$

So A takes 30 days and B takes 60 days. Their combined rate is:

$$\frac{1}{30}+\frac{1}{60}=\frac{1}{20}$$

Together they take 20 days.

Case 2: One worker is four times as efficient

Example: A is four times as efficient as B and finishes 60 days earlier. Find their combined time.

The efficiency ratio is $4:1$, so the time ratio is $1:4$. Let A take $x$ days and B take $4x$ days. The difference is:

$$4x-x=60$$ $$3x=\mathrm{60,}x=20$$

A takes 20 days and B takes 80 days. Their combined rate is:

$$\frac{1}{20}+\frac{1}{80}=\frac{5}{80}=\frac{1}{16}$$

Together they take 16 days.

Case 3: Percentage increase in efficiency

Example: A takes 24 days. B is 20 percent more efficient than A. Find B's time.

Let A's efficiency be 100 units. Twenty percent more makes B's efficiency 120 units. Their efficiency ratio is:

$$A:B=100:120=5:6$$

Reverse it for the time ratio:

$$T_A:T_B=6:5$$

If 6 parts of time equal 24 days, one part is 4 days. B takes 5 parts:

$$T_B=5\times 4=20\text{days}$$

B takes 20 days. A common mistake is reducing 24 directly by 20 percent. The correct decrease is not 20 percent because time is the reciprocal of efficiency.

More Workers Mean Less Time

If all workers have equal efficiency and work the same hours, the total effort can be measured in worker-days.

One worker-day means one worker working for one day. For a fixed job:

$$\text{Workers}\times \text{Days}=\text{constant}$$

This rule assumes equal worker efficiency and equal daily working hours.

Case 1: Change the number of workers

Example: Six workers finish a job in 20 days. How long will 10 workers take?

The total effort is:

$$6\times 20=120\text{worker-days}$$

Ten workers supply 10 worker-days of effort each day. The days needed are:

$$\frac{120}{10}=12\text{days}$$

Ten workers take 12 days. The answer is smaller than 20 because more workers share the same job.

Case 2: Workers needed for remaining work

Example: Five workers can finish a job in 18 days. They work for 12 days. How many workers are needed to finish the rest in 10 days?

The full job needs:

$$5\times 18=90\text{worker-days}$$

The first 12 days use:

$$5\times 12=60\text{worker-days}$$

Remaining work is $90-60=30$ worker-days. If $x$ workers must finish it in 10 days:

$$10x=30$$ $$x=3$$

Three workers are needed for the remaining 10 days.

Case 3: Deadline changes

Example: Twenty workers were assigned a project for 30 days. After 12 days, the remaining work must be finished in 10 days. Find the number of workers required from that point.

Total work is:

$$20\times 30=600\text{worker-days}$$

Work already completed is:

$$20\times 12=240\text{worker-days}$$

Remaining work is $600-240=360$ worker-days. To finish it in 10 days:

$$x\times 10=360$$ $$x=36$$

The project now requires 36 workers in total. Compared with the original 20 workers, this means $36-20=16$ extra workers are needed.

Workers, Days, Hours and Quantity of Work

When daily hours change, worker-days alone are not enough. Use worker-hours:

$$\text{Work}\propto \text{Workers}\times \text{Days}\times \text{Hours per day}$$

If the amount of output also changes, compare total effort per output unit.

Case 1: Find days when workers, hours and output change

Example: Eighteen workers working 6 hours daily complete 24 units in 8 days. How many days will 12 workers working 9 hours daily need for 36 units?

The first group uses:

$$18\times 6\times 8=864\text{worker-hours}$$

Those 864 worker-hours make 24 units. Therefore, one unit needs:

$$\frac{864}{24}=36\text{worker-hours}$$

Thirty-six units need:

$$36\times 36=1296\text{worker-hours}$$

The second group supplies this many worker-hours per day:

$$12\times 9=108$$

Therefore:

$$\text{Days}=\frac{1296}{108}=12$$

The second group needs 12 days.

Case 2: Find required daily hours

Example: Fifteen workers complete a job in 12 days at 8 hours per day. How many hours daily should 20 workers work to finish the same job in 18 days?

The first schedule needs:

$$15\times 12\times 8=1440\text{worker-hours}$$

Let the new daily hours be $h$. The second schedule must provide the same total effort:

$$20\times 18\times h=1440$$ $$360h=1440$$ $$h=4$$

The 20 workers need to work 4 hours per day. More workers and more days allow fewer hours each day.

Case 3: Machine-hours and production

Example: Five machines produce 300 units in 6 hours. How many units will 8 machines produce in 5 hours at the same rate?

The first case uses $5\times 6=30$ machine-hours. These 30 machine-hours produce 300 units, so one machine-hour produces:

$$\frac{300}{30}=10\text{units}$$

The second case has $8\times 5=40$ machine-hours. Therefore:

$$\text{Production}=40\times 10=400\text{units}$$

The machines produce 400 units.

Converting Different Worker Types into One Unit

Men, women and children may have different stated efficiencies. We cannot add their headcounts directly unless they are equally efficient. Convert everyone into one common worker unit first.

Example: Three men and five women complete a job in 18 days. One man does as much work as two women. How long will six men and ten women take?

Convert the first team into women. Three men equal six women, so:

$$3\text{men}+5\text{women}=6\text{women}+5\text{women}=11\text{women-equivalents}$$

For the second team, six men equal twelve women:

$$6\text{men}+10\text{women}=12+10=22\text{women-equivalents}$$

The second team has twice the effective workforce. For the same work, twice the rate needs half the time:

$$\text{New time}=\frac{18}{2}=9\text{days}$$

The second group takes 9 days.

Dividing Wages According to Work

Fair wages follow the work contributed, not merely the number of people. Work contribution equals efficiency multiplied by the time actually worked.

$$\text{Wage ratio}=\text{Work ratio}$$

Case 1: Everyone works for the same time

Example: A, B and C earn ₹6,600 for a job. Alone they take 10, 15 and 30 days. All three work together for the same time. Find C's share.

Because the time worked is equal, their work ratio equals their rate ratio:

$$\frac{1}{10}:\frac{1}{15}:\frac{1}{30}$$

Multiply each term by 30 to remove denominators:

$$A:B:C=3:2:1$$

There are $3+2+1=6$ total parts. C contributed 1 part:

$$C\text{'s share}=\frac{1}{6}\times 6,600=\text{Rs.}1,100$$

C receives ₹1,100.

Case 2: People work for different times

Example: A can finish a job in 20 days and works for 10 days. B can finish it in 30 days and works for 15 days. Total wages are ₹5,400. Find B's share.

A completes:

$$10\times \frac{1}{20}=\frac{1}{2}$$

B completes:

$$15\times \frac{1}{30}=\frac{1}{2}$$

They contribute equal work even though their individual speeds and working days differ. The wages are therefore divided equally:

$$B\text{'s share}=\frac{5,400}{2}=\text{Rs.}2,700$$

B receives ₹2,700.

Opposing Work: Building and Undoing

Not every worker helps. If one person builds while another destroys, their rates move in opposite directions. Add helpful rates and subtract opposing rates.

Example: A builds a wall in 12 days. B can undo the same wall in 36 days. How long will the wall take when both act together?

A adds this part per day:

$$\frac{1}{12}$$

B removes this part per day:

$$\frac{1}{36}$$

The actual forward progress is:

$$\text{Net rate}=\frac{1}{12}-\frac{1}{36}=\frac{3-1}{36}=\frac{1}{18}$$

Only one-eighteenth of the wall remains completed after both effects in one day. Therefore, the wall takes 18 days.

If the destructive rate were equal to or larger than the building rate, the wall would never be completed while both continued.

Pipes and Tanks Use the Same Work Model

A full tank is treated as one complete work. An inlet adds water, so its rate is positive. An outlet or leak removes water, so its rate is negative.

Case 1: Find a leak's emptying time

Example: A tap fills a tank in 8 hours, but with a leak the tank fills in 12 hours. Find how long the leak alone would take to empty the tank.

The tap tries to add one-eighth of the tank per hour:

$$\text{Tap rate}=\frac{1}{8}$$

With the leak, the actual net filling is only one-twelfth per hour:

$$\text{Net rate}=\frac{1}{12}$$

The missing part is what the leak removes:

$$\text{Leak rate}=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$$

The leak removes one twenty-fourth of the tank each hour, so it empties a full tank in 24 hours.

Case 2: Two inlets and one outlet

Example: Two inlets fill a tank in 6 and 8 hours. An outlet empties it in 24 hours. Find the time when all are open.

Add the inlet rates and subtract the outlet rate:

$$\text{Net rate}=\frac{1}{6}+\frac{1}{8}-\frac{1}{24}$$

Using 24 as the common denominator:

$$\text{Net rate}=\frac{4}{24}+\frac{3}{24}-\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$$

One-fourth of the tank fills each hour, so the full tank takes 4 hours.

Case 3: One pipe closes partway

Example: Pipe A fills a tank in 10 hours and pipe B in 15 hours. Both run for 4 hours, then B is closed. Find the total time from the start.

Together, their one-hour filling rate is:

$$\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$$

In the first 4 hours, they fill:

$$4\times \frac{1}{6}=\frac{2}{3}$$

One-third remains. After B closes, only A works at one-tenth per hour:

$$\text{Time for remainder}=\frac{\frac{1}{3}}{\frac{1}{10}}=\frac{10}{3}\text{hours}$$

We need the time measured from the start, so we also include the first 4 hours:

$$\text{Total time}=4+\frac{10}{3}=\frac{22}{3}=7\frac{1}{3}\text{hours}$$

The tank fills in $7\frac{1}{3}$ hours from the start.

Working Days, Breaks and Calendar Days

A working day is a day on which progress is made. A calendar day counts every day that passes, including breaks. A break changes elapsed time but does not change the amount of work.

Example: A needs 16 working days for a job. A works 4 days, takes a 2-day break and then resumes. Find the total calendar days.

A has already used 4 of the required 16 working days. The work rate does not change, so 12 working days remain:

$$16-4=12\text{working days}$$

The calendar includes the first work period, the break and the later work period:

$$4+2+12=18\text{calendar days}$$

The job finishes after 18 calendar days. The break is counted in time but contributes zero work.

Alternate-Day and Repeating Schedules

When workers follow a repeating schedule, calculate the work done in one full cycle. Repeat complete cycles without changing their order, then handle the last partial cycle carefully.

Example: A can finish a job in 12 days and B in 18 days. They work on alternate days, beginning with A. Find when the job finishes.

A's daily work is $\frac{1}{12}$ and B's is $\frac{1}{18}$. One two-day cycle completes:

$$\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$$

Seven complete cycles use 14 days and complete:

$$7\times \frac{5}{36}=\frac{35}{36}$$

Only one thirty-sixth remains. Day 15 belongs to A, whose full-day rate is one-twelfth, or three thirty-sixths. A needs only one-third of that day to finish one thirty-sixth:

$$\frac{\frac{1}{36}}{\frac{1}{12}}=\frac{1}{3}\text{day}$$

The work finishes in $14\frac{1}{3}$ days. We cannot simply use the average two-day rate for the final part without checking whose turn comes next.

Data Sufficiency in Time and Work

Data sufficiency checks whether the given information leads to one definite result. Check each statement alone before combining them.

Example: Can B finish a job in less than 30 days?

• Statement I: A alone takes 30 days.
• Statement II: A and B together take 20 days.

Statement I gives only A's rate. It tells us nothing about B, so it is not sufficient alone.

Statement II gives only the pair's rate. Without A's rate, B cannot be separated, so it is also not sufficient alone.

Together:

$$B\text{'s rate}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$$

B takes 60 days, so the stated time is not below 50 days. Both statements together are sufficient, while neither is sufficient alone.

Common Mistakes and Why They Are Wrong

Adding times instead of rates

People working together add the work they complete each day. Their individual completion times are not added. A 10-day worker and a 15-day worker do not take 25 days together.

Forgetting to take the reciprocal

If a rate is $\frac{5}{36}$ of a job per day, the time is $\frac{1}{\frac{5}{36}}=\frac{36}{5}$ days. The rate itself is not the time.

Using direct proportion for workers and days

For the same job, more equal workers mean fewer days. Workers and days are inversely related, so doubling workers halves time.

Subtracting a percentage directly from time

If efficiency rises by 20 percent, time does not automatically fall by 20 percent. New time is old time divided by 1.2 because time is inversely related to efficiency.

Forgetting completed work before a change

When a person leaves or a pipe closes, do not apply the new rate to the whole job. First find the work already completed, then apply the new rate only to what remains.

Adding an outlet or destructive worker

An outlet removes water and a destructive worker undoes work. Their rates must be subtracted from helpful rates.

Dividing wages by time only

Wages follow actual work, which depends on both efficiency and time worked. A slower worker may work longer and still contribute the same amount.

Formulas and Cheat Sheet

Situation	Formula or decision	Why it works
Basic relationship	Work $=$ Rate $\times$ Time	Equal rate is repeated for each time unit
One-day work	$\frac{1}{D}$	One whole job is shared across $D$ days
Time for required work	$\frac{\text{Work}}{\text{Rate}}$	Count how many daily-rate pieces make the work
People helping together	Add their rates	Each person adds completed work
One person opposes another	Subtract the opposing rate	Opposing work removes progress
Remaining work	$1-$ completed work	The full job is one whole
Equal workers, fixed hours	Workers $\times$ Days $=$ constant	Total effort stays fixed
Workers and daily hours	Workers $\times$ Days $\times$ Hours	Measures total worker-hours
Efficiency and time	$E_1{T}_1=E_2{T}_2$	Same work equals rate times time
Wages	Wage ratio $=$ work ratio	Payment follows contribution
Inlets and outlets	Add inlets, subtract outlets	Inlets fill while outlets empty
Repeating schedule	Find one cycle's work	The same rate pattern repeats

Safe Shortcuts

• Choose total work as the LCM of the individual times. Dividing that work by each time gives whole-number daily efficiencies and often removes fractions.
• If A and B alone take $a$ and $b$ days and both help throughout, their combined time is $\frac{ab}{a+b}$. Do not use this when one person works against the other or leaves partway.
• Reverse an efficiency ratio to get the time ratio for the same job. If efficiencies are $m:n$, individual times are $n:m$.
• For equal workers with unchanged daily hours, use workers $\times$ days $=$ constant. Include hours as another factor when daily working hours change.
• To find a missing worker's rate, subtract the known worker rates from the full team's rate, then take the reciprocal to find individual time.
• Divide wages in the ratio rate $\times$ time worked. Use the rate ratio alone only when everyone works for the same time.
• For pipes and opposing work, add filling or building rates and subtract emptying or undoing rates before taking the reciprocal.

These shortcuts are safe only when the job, work periods and direction of every rate are clear. They should shorten a correct work-rate model, not replace it.

Frequently Asked Questions

What is one-day work?

One-day work is the fraction of the whole job completed in one day. If a person takes 20 days for the full job, the one-day work is $\frac{1}{20}$.

Why is total work taken as 1?

The exact physical size of the job is often not given or needed. Calling the complete task 1 lets every person's contribution be written as a fraction of the same whole.

Why are rates added when people work together?

Each person completes a part of the same job during the day. The total completed part is the sum of those individual parts.

Why are efficiency and time inversely proportional?

For the same work, a person who completes more work each day needs fewer days. Their product stays equal to the fixed total work.

What is a worker-day?

One worker-day is the effort of one worker working for one day. It is useful only when the workers have equal efficiency and daily hours are unchanged.

When should worker-hours be used?

Use worker-hours when the number of hours worked per day changes. Total effort then depends on workers, days and hours per day.

What happens when someone leaves before the work finishes?

Find the work completed before the person leaves. Subtract it from the whole job. Then use the remaining team's rate only on the remaining work.

How are wages divided when people complete different amounts of work?

Divide wages in the ratio of work done. If everyone works for the same time, this is the efficiency ratio. If their times differ, calculate efficiency multiplied by time for each person.

Are pipes and tanks different from Time and Work?

They use the same mathematics. Filling is positive work and emptying is negative work. A full tank is treated as one complete job.

What is the fastest safe method for combined work?

Use either fraction rates or choose LCM work units. The LCM method often makes arithmetic easier, but both methods must begin with each person's one-time-unit work.

Summary

The whole topic comes from one relationship: work equals rate multiplied by time. When a full job takes $D$ days, one day's work is $\frac{1}{D}$. Helping rates are added, opposing rates are subtracted, and time is found by dividing the required work by the available rate.

For changing schedules, always separate the stages. Find what is completed before someone joins, leaves, rests or closes a pipe. Then calculate the remaining work with the new rate. For groups, machines and changing hours, measure the fixed effort with worker-days, worker-hours or machine-hours.

Most shortcuts are only shorter versions of this reasoning. Once the whole work, each one-day rate, and the reason for adding or subtracting rates are clear, every later situation becomes another use of the same idea.