Whole work
The complete job may be set equal to \(1\) or to a convenient LCM number of units. Both models are equivalent.
10 SSC CGL · Mathematics
Time and Work
Handwritten-style notes for SSC CGL Tier I and Tier II.
01MAP IT!
The Big Picture — Graphic Mind Map
Convert every worker, machine or pipe into a signed work rate; then add rate × active time across the schedule.
TIME
RATE
WORK
RATE
WORK
1. Work Fractionwhole job, completed part and remaining part
2. Efficiencywork per unit time; inverse of time
3. Unit WorkLCM-based integer capacities
4. Changing Teamsjoin, leave, alternate and partial schedules
5. Work & Wagespay in proportion to actual contribution
6. Pipesinlets positive; outlets and leaks negative
One engine drives the chapter
Choose a whole-work scale. Rate tells how much is done per unit time. Across multiple stages, add each stage’s signed contribution.
\[\boxed{W=RT}\qquad\boxed{W_{\text{total}}=\sum R_it_i}\]
\[\text{worker/inlet rate}\gt0,\qquad\text{outlet/leak rate}\lt0\]
02ZERO TO EXPERT
The Foundation (Basics) — Complete Concept Build
Begin with one whole job, learn reciprocal rates, then scale to teams and signed pipe systems.
AWork, rate, time and efficiency
Rate \(R\)
Work completed per unit time.
\[R=\frac WT\]
Time \(T\)
\[T=\frac WR\]
For a fixed job, higher rate means less time.
One-person rate
If \(A\) completes the whole job in \(a\) days:
\[R_A=\frac1a\text{ job/day}\]
Work in \(t\) days
\[W_A(t)=\frac ta\]
Remaining work is \(1-t/a\).
Efficiency
Efficiency is proportional to work rate.
\[E\propto R\qquad\text{and for fixed work}\qquad E\propto\frac1T\]
Time and efficiency run in opposite directions
For equal work, if time ratio is \(a:b\), efficiency ratio is reversed.
\[T_A:T_B=a:b\Rightarrow E_A:E_B=b:a\]
BFraction method vs LCM/unit-work method
Fraction method
Set total work \(W=1\). A worker taking \(a\) days has rate \(1/a\). Best for compact algebra.
LCM method
Set total work to the LCM of individual times. Every rate becomes an integer number of units per day.
\[W=\operatorname{LCM}(a,b,c)\]
Integer efficiencies
\[E_A=\frac Wa,\qquad E_B=\frac Wb\]
The units are artificial but the ratio is exact.
The LCM turns fractions into bricks
If \(A\) and \(B\) take \(12\) and \(18\) days, choose \(36\) work units. Their daily capacities become \(3\) and \(2\) units.
\[36/12=3,\qquad36/18=2\]
CCombined work and efficiency ratios
Same-direction rates add
\[R_{A+B}=\frac1a+\frac1b\]
\[T_{A+B}=\frac{ab}{a+b}\]
Three workers
\[T_{A+B+C}=\frac{abc}{ab+bc+ca}\]
Difference of workers
If one undoes the other:
\[R_{\text{net}}=\left|\frac1a-\frac1b\right|\]
Together time and one alone known
If \(A+B\) finish in \(x\) days and \(A\) alone in \(a\) days, with \(a\gt x\):
\[T_B=\frac{ax}{a-x}\]
Efficiency ratio to time ratio
\[E_A:E_B=m:n\Rightarrow T_A:T_B=n:m\]
Known combined efficiency
\[E_A+E_B=E_{A+B}\]
Use signed efficiency if one agent destroys work.
Workers combine like meshing gears
Count work completed per day, not days themselves. Add daily rates, then invert only at the end to get time.
\[\text{combined time}=\frac1{\text{sum of rates}}\]
DComparative statements about speed
\(x\%\) more efficient
\[E_A=E_B\left(1+\frac{x}{100}\right)\]
\[\frac{T_A}{T_B}=\frac{100}{100+x}\]
\(x\%\) less efficient
\[E_A=E_B\left(1-\frac{x}{100}\right)\]
\[\frac{T_A}{T_B}=\frac{100}{100-x}\]
Difference in completion times
If efficiency ratio \(m:n\) gives time ratio \(n:m\) and time difference is \(D\), one ratio unit is \(D/|n-m|\).
EMen–days–hours and work equivalence
Homogeneous workforce
\[W\propto M\times D\times H\times E\]
\(M\) workers, \(D\) days, \(H\) hours per day, efficiency \(E\).
Same work comparison
\[M_1D_1H_1E_1=M_2D_2H_2E_2\]
Different work quantities
\[\frac{M_1D_1H_1E_1}{W_1}=\frac{M_2D_2H_2E_2}{W_2}\]
Four levers produce output
More workers, more days, more hours or more individual efficiency raise work directly—provided productivity remains uniform.
\[W=kMDHE\]
Men and women or skilled groups
Convert unlike workers using their efficiency ratio. If \(E_M:E_W=a:b\), then one man is equivalent to \(a/b\) women.
Uniformity condition
The direct-product model assumes identical daily hours, constant individual efficiency and no crowding or learning effect unless stated.
FPartial work, joining and leaving
Stage method
\[W_1=R_1t_1,\quad W_2=R_2t_2,\quad\ldots,\quad\sum W_i=1\]
Remaining work
\[W_{\text{left}}=1-W_{\text{done}}\]
Divide remaining work by the next-stage rate.
Join or leave
At the change time, close the first stage. Build a new combined rate for the next interval.
Changing teams create a rate timeline
Do not average team sizes. Calculate work in each interval using the team active during that interval.
\[R_1t_1+R_2t_2+R_3t_3=1\]
GAlternate days and cyclic schedules
Two-day cycle
If \(A\) works first and \(B\) second:
\[W_{\text{cycle}}=R_A+R_B\]
Complete cycles first
\[k=\left\lfloor\frac{1}{W_{\text{cycle}}}\right\rfloor\]
Use the largest number of full cycles that does not finish or exceed the job, then simulate the residual days in order.
Starter matters
If the work finishes on an odd day, the person who starts the cycle may perform the last partial day. Reversing order can change completion time.
Alternate work is a repeating calendar strip
Box the shortest repeating cycle. Multiply complete cycles, then inspect the remaining work one scheduled day at a time.
\[\text{total time}=\text{full-cycle time}+\text{residual time}\]
HWork and wages
Contribution rule
\[\text{wage share}\propto\text{actual work}=E\times t\]
Wage ratio
\[W_A:W_B=E_At_A:E_Bt_B\]
Equal working time
\[t_A=t_B\Rightarrow W_A:W_B=E_A:E_B\]
Divide money by completed work, not headcount
If two workers differ in efficiency or active time, equal wages are generally wrong. Form each worker’s work contribution first.
IPipes and cisterns — signed work rates
Inlet
An inlet filling a tank in \(a\) hours has positive rate:
\[R_{\text{in}}=+\frac1a\]
Outlet or leak
An outlet emptying a full tank in \(b\) hours has negative rate:
\[R_{\text{out}}=-\frac1b\]
Net pipe rate
\[R_{\text{net}}=\sum R_{\text{in}}-\sum|R_{\text{out}}|\]
\[T=\frac1{R_{\text{net}}}\]
Water enters with plus and leaves with minus
The tank is simply a “whole job.” A net positive rate fills it, a net negative rate empties it, and zero net rate leaves the level unchanged.
\[R_{\text{net}}\gt0:\text{ fill}\quad R_{\text{net}}\lt0:\text{ empty}\quad R_{\text{net}}=0:\text{ steady}\]
JPipe formulas and leak discovery
One inlet, one outlet
If inlet fills in \(a\) hours and outlet empties in \(b\) hours, with \(b\gt a\):
\[T=\frac{ab}{b-a}\]
Two inlets
\[T=\frac{ab}{a+b}\]
Leak’s emptying time
If inlet alone fills in \(a\) hours but with a leak fills in \(b\) hours, where \(b\gt a\):
\[T_{\text{leak}}=\frac{ab}{b-a}\]
Net emptying case
If outlet is faster than inlet, the full tank empties in
\[T=\frac{ab}{a-b}\quad(a\gt b)\]
Partially filled tank
If initial filled fraction is \(f\) and net filling rate is \(R\gt0\):
\[T=\frac{1-f}{R}\]
Partially full, net emptying
For net emptying magnitude \(|R|\):
\[T=\frac{f}{|R|}\]
A leak is the missing negative rate
Subtract the observed net filling rate from the inlet’s normal filling rate.
\[R_{\text{leak}}=\frac1a-\frac1b\]
KPipes opened or closed at different times
Stage the schedule
\[\sum R_it_i=\text{required tank fraction}\]
Outlet opened late
First calculate the fraction filled by inlets alone. For the remaining fraction, use the new signed net rate.
Overflow or delayed closure
Once the tank reaches full capacity, additional positive work cannot be stored unless overflow is explicitly part of the question.
A scheduled tap is a rate timeline
Every opening or closing event creates a new interval. The method is identical to workers joining or leaving.
\[R_1t_1+R_2t_2+\cdots=1-f_0\]
LDecision table and edge conditions
| Pattern | Model | Critical condition |
|---|---|---|
| One agent’s completion time | Rate \(=1/T\) | Whole job is fixed |
| Several agents together | Add same-direction rates | They work simultaneously and independently |
| Men–days–hours | \(W\propto MDHE\) | Efficiency and hours are uniform unless stated |
| Joining or leaving | Stage-wise rate timeline | Close each interval at the exact change point |
| Alternate work | Cycle plus residual days | Starting order matters |
| Wages | Share by actual work | Use efficiency × active time |
| Inlet and outlet | Signed rates | Net direction decides filling or emptying |
| Leak discovery | Normal rate minus observed net rate | Observed filling time must exceed normal filling time |
03SAVE TIME
Short Tricks & Magic Formulas
Choose integer work units, add signed efficiencies, and postpone division until the final step.
1Two-agent speed formulas
Both complete
\[a,b\Rightarrow T=\frac{ab}{a+b}\]
One undoes
\[a,b\Rightarrow T=\frac{ab}{|b-a|}\]
Find missing worker
\[T_{A+B}=x,\ T_A=a\Rightarrow T_B=\frac{ax}{a-x}\]
2Efficiency conversions
Reverse the ratio
\[E_A:E_B=m:n\Rightarrow T_A:T_B=n:m\]
More efficient to time
\[x\%\text{ more efficient}\Rightarrow T\text{ factor}=\frac{100}{100+x}\]
Less efficient to time
\[x\%\text{ less efficient}\Rightarrow T\text{ factor}=\frac{100}{100-x}\]
3LCM and partial-work shortcuts
LCM total
\[W=\operatorname{LCM}(T_1,T_2,\ldots)\]
Remaining time
\[t_{\text{left}}=\frac{W-W_{\text{done}}}{R_{\text{new}}}\]
Fraction after \(t\) days
\[\text{done}=t\sum R_i\]
4Men–days–hours and wages
Same work cross-product
\[M_1D_1H_1E_1=M_2D_2H_2E_2\]
Wage ratio
\[W_A:W_B=E_At_A:E_Bt_B\]
Team replacement
Convert each worker type to one common efficiency unit before comparing team sizes.
5Alternate days and pipe hacks
Cycle first
\[W_{\text{cycle}}=\sum R_{\text{scheduled days}}\]
Use full cycles, then check remaining days in sequence.
Leak shortcut
\[a\text{ alone},\ b\text{ with leak}\Rightarrow T_L=\frac{ab}{b-a}\]
Partial tank shortcut
\[T=\frac{\text{required change in tank fraction}}{|R_{\text{net}}|}\]
SSC time-work speed dashboard
Choose total work, write signed rates, draw the schedule, add completed units, and divide the remaining work by the final rate.
\[\text{whole}\to\text{rates}\to\text{timeline}\to\text{done}\to\text{remaining}\]
04DON'T RUSH
The SSC / TCS Traps — Red Flags 🚩
Wrong options add completion times, forget inverse efficiency, or treat outlets as positive rates.
Trap 1: times added instead of rates
- Workers combine through work per day, not days per job.
- Compute \(1/a+1/b\), then invert.
- Combined time for positive workers must be less than each individual time.
Trap 2: efficiency and time ratios not reversed
- For fixed work, greater efficiency means smaller time.
- Reverse the ratio: \(E_A:E_B=m:n\Rightarrow T_A:T_B=n:m\).
- “\(25\%\) faster” does not mean “\(25\%\) less time.”
Trap 3: outlet/leak sign ignored
- Inlets are positive; outlets and leaks are negative.
- If net rate is zero, the level never changes.
- If net rate is negative, an empty tank cannot continue “emptying”; initial level matters.
Average team size used
When people join or leave, compute work in each interval. An average headcount generally ignores duration and efficiency.
Alternate order ignored
Complete cycles may match, but the starting worker determines the residual day and can change the answer.
Wages divided equally
Wages follow actual work \(Et\), not number of workers, unless contributions are equal.
05LOCK IT IN
Memory Hooks & Mnemonics
Remember rates as taps, gears and daily work bricks.
“Work sits above rate × time”
Cover the unknown in the triangle: multiply across the base or divide from the top.
“Efficiency up, time down”
For the same job, reverse efficiency and completion-time ratios.
“LCM turns days into daily bricks”
Choose LCM as total work; divide by each time to obtain integer daily capacity.
“New person, new rate”
Every joining, leaving, opening or closing event begins a new stage on the timeline.
“Fill is plus; drain is minus”
Treat every pipe as a signed worker on the tank job.
“Money follows the work slice”
Divide wages in the ratio of efficiency × active time.
One-person rate\(1/T\)
Same-direction teamadd rates
Efficiency and timeinverse ratios
Changing schedulesum stage work
Outlet or leaknegative rate
Wagesproportional to \(Et\)