Time
The duration needed to complete a stated quantity of work.
S4 · Topic 10 · Reasoning & Quant
Time & Work, Pipes & Cisterns
Notes for UPSC Principal & Vice-Principal examinations only.
01Basics
The Foundation — Work Is a Rate Problem
Convert every worker or pipe into a fraction of the total task completed per unit time.
AThe four building blocks
Work
A complete job, chosen as \(1\) whole or a convenient number of units.
Rate
Work completed per unit time: fraction per day, units per hour, or tank per minute.
Efficiency
The speed of doing work. Greater efficiency means less time for the same task.
BThe work–rate–time triangle
Universal relationship\[W=R\times T,\qquad R=\frac WT,\qquad T=\frac WR\]
- If total work is taken as \(1\), a person finishing in \(x\) days has rate \(\frac1x\) work per day.
- Rates of simultaneous helpers are added.
- Rates of destroyers, leaks or outlets are subtracted.
- All rates must use the same time unit.
Unit discipline
A daily rate cannot be added to an hourly rate. Convert first: if a worker completes \(\frac1{12}\) of a job per day in an \(8\)-hour day, the hourly rate is \(\frac1{96}\).
CTwo valid models: unit fraction and LCM work
Model 1 — Total work \(=1\)
If A finishes in \(12\) days and B in \(18\) days:
\[R_A=\frac1{12},\quad R_B=\frac1{18}\]\[R_{A+B}=\frac1{12}+\frac1{18}=\frac5{36}\]
\[T_{A+B}=\frac1{5/36}=\frac{36}{5}=7\frac15\text{ days}\]
Best when fractions remain simple.
Model 2 — LCM total work
Take total work as \(\operatorname{LCM}(12,18)=36\) units.
\[E_A=\frac{36}{12}=3\text{ units/day},\quad E_B=\frac{36}{18}=2\text{ units/day}\]
\[T_{A+B}=\frac{36}{3+2}=\frac{36}{5}\text{ days}\]
Best for joins, leaves, alternate days and wages.
1. List times\(12,18\)
2. Take LCM\(36\) units
3. Find rates\(3,2\)
4. Apply scheduleAdd/subtract units
5. DivideWork ÷ rate
DEfficiency and time are inversely proportional
Inverse relation
\[E\propto\frac1T,\qquad E_1T_1=E_2T_2\]
For the same work and working conditions, doubling efficiency halves time.
Efficiency ratio from time
\[E_A:E_B=T_B:T_A\]
If A takes \(12\) days and B takes \(18\) days, \(E_A:E_B=18:12=3:2\).
Time after efficiency change
\[T_2=T_1\frac{E_1}{E_2}\]
A \(25\%\) increase in efficiency gives \(T_2=T_1/1.25=0.8T_1\), a \(20\%\) time reduction.
ECore team formulas
| Situation | Combined rate | Time / inference |
|---|---|---|
| A takes \(x\), B takes \(y\) | \(\frac1x+\frac1y=\frac{x+y}{xy}\) | \(T=\frac{xy}{x+y}\) |
| A works; B destroys | \(\frac1x-\frac1y\), assuming \(x\lt y\) | \(T=\frac{xy}{y-x}\) |
| A+B take \(t\); A takes \(x\) | \(R_B=\frac1t-\frac1x\) | \(T_B=\frac{xt}{x-t}\) |
| Efficiency ratio \(a:b\) | Rates are \(a:b\) | Times are \(b:a\) |
| Same work, groups | \(M_1D_1E_1=M_2D_2E_2\) | When daily hours are equal. |
| Hours also vary | \(M_1D_1H_1E_1=M_2D_2H_2E_2\) | For the same quantity of work. |
Combined work
A can finish a report in \(15\) days and B in \(10\) days.
\[T=\frac{15\times10}{15+10}=6\text{ days}\]
Together they take \(6\) days.
Find the unknown worker
A+B take \(8\) days; A alone takes \(12\) days.
\[R_B=\frac18-\frac1{12}=\frac1{24}\]
B alone takes \(24\) days.
Different efficiencies
A is \(50\%\) more efficient than B. If B takes \(18\) days:
\[E_A:E_B=3:2\Rightarrow T_A:T_B=2:3\]
A takes \(12\) days.
FPipes and cisterns: the same mathematics with signs
Filling pipePositive rate
Emptying pipeNegative rate
Full tankWork \(=1\)
Net tank rate\[R_{\text{net}}=\sum R_{\text{inlets}}-\sum R_{\text{outlets}}\]
If the net rate is positive, the tank fills. If it is negative, the tank empties. If it is zero, the level remains unchanged.
Two inlet pipes
Pipes fill in \(12\) and \(18\) hours.
\[T=\frac{12\times18}{12+18}=\frac{36}{5}=7\frac15\text{ h}\]
Both fill the tank in \(7\) h \(12\) min.
Inlet and outlet
An inlet fills in \(8\) h; an outlet empties in \(12\) h.
\[R=\frac18-\frac1{12}=\frac1{24}\]
With both open, the tank fills in \(24\) h.
Capacity form
A \(600\)-litre tank receives \(50\) L/min and loses \(20\) L/min.
\[T=\frac{600}{50-20}=20\text{ min}\]
Net discharge is \(30\) L/min.
02Expert
The Deep Dive — Schedules, Groups and Variable Rates
Master join–leave patterns, alternate work, workforce equivalence, leaks and repeated cycles.
AWhen workers join or leave
Phase method
Break the schedule whenever the active team changes.
\[W_1=R_1T_1,\qquad W_{\text{left}}=W-W_1\]\[T_2=\frac{W_{\text{left}}}{R_2}\]
Never average the two rates unless the durations are equal and the question permits it.
A leaves after some days
A takes \(12\) days and B \(18\) days. They work together for \(4\) days; then A leaves.
\[W_1=4\left(\frac1{12}+\frac1{18}\right)=\frac59\]\[W_{\text{left}}=\frac49,\quad T_B=\frac{4/9}{1/18}=8\text{ days}\]
Total elapsed time: \(12\) days.
StartA + B begin
Phase 1Use \(R_A+R_B\)
Change pointA leaves
Phase 2Use \(R_B\)
FinishAdd phase times
BAlternate-day and repeating schedules
Cycle principle
If A and B work on alternate days, one complete \(2\)-day cycle produces:
\[W_{\text{cycle}}=R_A+R_B\]
Count full cycles, then assign the residual work to the person whose turn comes next.
Alternate days
A takes \(10\) days, B \(15\) days; A starts.
\[W_{2\text{ days}}=\frac1{10}+\frac1{15}=\frac16\]
Five cycles complete \(\frac56\) in \(10\) days. A completes the last \(\frac16\) in \(\frac{5}{3}\) days.
Total: \(11\frac23\) days.
Last partial day
A “day” in the answer may be fractional unless the wording says work is performed only in whole-day shifts. Do not round automatically.
\[T_{\text{last}}=\frac{W_{\text{remaining}}}{R_{\text{next}}}\]
CPairwise team data
Given pair rates
Let individual rates be \(a,b,c\). Then:
\[a+b=p,\qquad b+c=q,\qquad c+a=r\]
\[a+b+c=\frac{p+q+r}{2}\]
Individual rates:
\[a=\frac{p+r-q}{2},\quad b=\frac{p+q-r}{2},\quad c=\frac{q+r-p}{2}\]
Three-team reconstruction
A+B take \(12\) days, B+C \(15\) days, and C+A \(20\) days.
\[a+b+c=\frac12\left(\frac1{12}+\frac1{15}+\frac1{20}\right)=\frac1{10}\]
All three together finish in \(10\) days.
DMen–days–hours–efficiency model
General production model\[W\propto M\times D\times H\times E\]\[\frac{W_1}{W_2}=\frac{M_1D_1H_1E_1}{M_2D_2H_2E_2}\]
| Quantity | Meaning | How it affects output |
|---|---|---|
| \(M\) | Number of workers or equivalent workers | Directly proportional, if individual efficiency is unchanged. |
| \(D\) | Number of working days | Directly proportional. |
| \(H\) | Working hours per day | Directly proportional within the assumed range. |
| \(E\) | Efficiency per worker per hour | Directly proportional. |
| \(W\) | Quantity of work | Do not cancel when the two jobs differ in size. |
Changed workforce and hours
\(24\) workers, \(8\) h/day, finish in \(15\) days. Days for \(30\) workers at \(6\) h/day:
\[24\times15\times8=30\times D\times6\]
\(D=16\) days.
More work too
If the new job is \(k\) times the old work:
\[\frac{M_1D_1H_1E_1}{M_2D_2H_2E_2}=\frac{W_1}{W_2}=\frac1k\]
Never use equal-work cancellation when \(k\ne1\).
Workforce reduction
If \(x\%\) of workers leave and output must remain fixed, time multiplies by:
\[\frac{100}{100-x}\]
A \(20\%\) workforce fall increases time by \(25\%\).
EEquivalent workers: men, women, machines or teams
Convert to one common unit
If \(2\) skilled workers do as much work as \(3\) trainees:
\[2S=3T\Rightarrow 1S=\frac32T\]
A mixed team of \(8S+12T\) equals \(12T+12T=24T\).
Efficiency from stated equivalence
If \(a\) workers of type X equal \(b\) workers of type Y:
\[aE_X=bE_Y\Rightarrow E_X:E_Y=b:a\]
The number ratio reverses to give the efficiency ratio.
FWork and wages
Basic rule
\[\text{Wage share}\propto\text{work done}=E\times T\]
Equal time means wages follow efficiency. Equal efficiency means wages follow time.
General wage ratio
\[W_A:W_B=E_AT_A:E_BT_B\]
If hours per day differ, use total hours or include them separately.
Unequal time
Efficiencies A:B are \(3:2\). A works \(4\) days and B \(6\) days.
\[W_A:W_B=(3\times4):(2\times6)=1:1\]
They receive equal shares.
GEfficiency change and time change are not equal percentages
Efficiency rises by \(x\%\)
\[\text{Time fall}=\frac{100x}{100+x}\%\]
For a \(20\%\) efficiency rise, time falls by \(\frac{2000}{120}=16\frac23\%\).
Efficiency falls by \(x\%\)
\[\text{Time rise}=\frac{100x}{100-x}\%\]
For a \(20\%\) efficiency fall, time rises by \(25\%\).
HLeaks, delayed outlets and observed-time problems
Find a hidden leak
An inlet alone fills in \(x\) hours, but with a leak it takes \(y\) hours where \(y\gt x\).
\[R_L=\frac1x-\frac1y\]
\[T_L=\frac{xy}{y-x}\]
Leak capacity
A pipe fills in \(6\) h; with a leak, it takes \(8\) h.
\[R_L=\frac16-\frac18=\frac1{24}\]
The leak alone empties the full tank in \(24\) h.
Outlet opened later
If the inlet works alone for \(t_1\), then an outlet opens:
\[W_1=\frac{t_1}{x},\quad W_{\text{left}}=1-W_1\]\[T_2=\frac{W_{\text{left}}}{1/x-1/y}\]
IRepeated opening cycles
Cycle-rate method
If inlet A runs for \(a\) minutes and outlet B for \(b\) minutes in every cycle:
\[W_{\text{cycle}}=\frac a{x}-\frac b{y},\qquad T_{\text{cycle}}=a+b\]
Count full cycles, then analyse the remaining fraction in the exact order of the next cycle.
Alternating pipe schedule
A fills in \(10\) h and B empties in \(15\) h. A runs for \(2\) h, then B for \(1\) h.
\[W_{3\text{ h cycle}}=\frac2{10}-\frac1{15}=\frac2{15}\]
Seven cycles fill \(\frac{14}{15}\) in \(21\) h. A then fills \(\frac1{15}\) in \(\frac23\) h.
Total: \(21\frac23\) h.
JTank volume and level
Uniform cross-section
Tank volume is proportional to water height.
\[\frac{V_1}{V_2}=\frac{h_1}{h_2}\]
Half the height means half the volume only for a uniform tank.
Known capacity
\[T=\frac{\text{volume to be added}}{\text{net discharge}}\]
Use litres/minute or cubic metres/hour consistently.
Partly filled tank
If fraction \(f\) is already filled, remaining work is:
\[W_{\text{left}}=1-f\]
Do not calculate from an empty tank unless stated.
03Accuracy
The UPSC Trap / Watch Out
Protect accuracy when signs, schedules, efficiencies and units change.
AClassic distractors and reasoning errors
Adding times
Workers working together add rates, not completion times.
Efficiency ratio not reversed
If time ratio is \(2:3\), efficiency ratio is \(3:2\).
Outlet treated as positive
An emptying pipe or leak has a negative rate while the tank is filling.
Mixed time units
Convert hours to minutes or days to hours before combining rates.
Changing team ignored
Whenever someone joins or leaves, close the current phase and start a new rate.
Wrong alternate-day starter
The final fraction depends on whose turn begins the cycle and whose turn comes next.
Whole-day rounding
Do not round a fractional finishing day unless work is restricted to completed shifts.
Workers assumed identical
Men–days cancellation requires equal or explicitly adjusted efficiencies.
Hours per day omitted
When daily working hours change, include \(H\) in the production equation.
Different work quantities cancelled
If one job is larger, retain the work ratio \(W_1:W_2\).
Wages divided by time only
Wages follow work contribution, which depends on both efficiency and duration.
Leak time subtracted directly
Subtract rates \(\frac1x-\frac1y\), not hours \(y-x\).
Initial tank level ignored
A partly filled tank requires only the unfilled fraction.
Net rate sign not checked
If outlet capacity equals or exceeds inlet capacity, the tank will not fill.
Percent efficiency equals percent time
Inverse variation makes the percentage changes unequal.
Average rate used casually
Average rates require correct time weights; phase calculation is safer.
BHigh-speed shortcuts and mental methods
Two helpers
For times \(x,y\), together time is \(\frac{xy}{x+y}\).
Helper and destroyer
For positive time \(x\) and negative time \(y\), net time is \(\frac{xy}{y-x}\), with \(y\gt x\).
LCM units
Choose total work as the LCM of individual times; rates become small integers.
Efficiency reversal
Write time ratio, reverse it immediately, then label it efficiency.
Pair-sum shortcut
Add the three pair rates and halve to obtain the rate of all three.
Percent conversion
Efficiency rise \(x\%\) gives time fall \(\frac{100x}{100+x}\%\).
Cycle block
Calculate work per full cycle first; handle only the residual separately.
Contribution chart
For wages, write \(E\times T\) under each worker, reduce the ratio, then divide money.
Direction check
Adding help must reduce time; adding a leak must increase filling time.
The safest exam sequence
Fix the total work, convert every agent to a signed rate, draw the schedule in phases, compute remaining work after each phase, and only then convert the final fraction into time.
04Revise
Last-Minute Master Sheet
Compact relationships for rapid revision.
Work–rate–time\[W=RT,\quad R=\frac WT,\quad T=\frac WR\]
One-person rate\[T=x\Rightarrow R=\frac1x\]
Two helpers\[T=\frac{xy}{x+y}\]
One helper, one destroyer\[T=\frac{xy}{y-x},\quad y\gt x\]
Efficiency and time\[E_1:E_2=T_2:T_1\]
Workforce model\[W\propto MDHE\]
Wage ratio\[W_A:W_B=E_AT_A:E_BT_B\]
All three from pairs\[a+b+c=\frac{(a+b)+(b+c)+(c+a)}2\]
Net pipe rate\[R_{\text{net}}=\sum R_{\text{in}}-\sum R_{\text{out}}\]
Leak alone\[T_L=\frac{xy}{y-x}\]
Cycle work\[W_c=\sum R_iT_i\]
Partly filled tank\[W_{\text{left}}=1-f\]
QQuestion-type map
Single schedule?Add signed rates.
Team changes?Split into phases.
Pattern repeats?Use cycle work.
People/hours vary?Use \(MDHE\).
One-line memory rule
Time is obtained only after the relevant work is divided by the net signed rate for that phase.