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Time, Work, and Distance
Work Rate Fundamentals
Every work problem boils down to a single relationship: the total work done equals the work rate multiplied by the time spent. If a job is treated as one whole unit, then the rate of a person is the fraction of the job they complete per unit time.
$$W = R \times T$$
Total work equals rate multiplied by time — the foundation of every work problem.
$W$=Total work done (often 1 for one complete job)(jobs or units)
$R$=Work rate — fraction of the job completed per unit time(jobs per unit time)
$T$=Time spent working(hours, days, etc.)
$W = 1$
→The rate is $R = \frac{1}{T}$, meaning if someone finishes in $T$ days, their daily rate is $\frac{1}{T}$.
Whole Job Convention: Set $W = 1$ for "one complete job". A person who finishes in 10 days has rate $R = \frac{1}{10}$ jobs/day.
Rate from Time: Given completion time $T$, the rate is always $R = \frac{1}{T}$. Shorter time means faster rate.
Work from Rate and Time: If you know the rate and time, multiply: $W = R \times T$. The answer is a fraction of the total job.
Combined Work and Efficiency
When two or more people work together, their rates add. This is because each person independently contributes their share of work per unit time.
$$R_{\text{total}} = R_1 + R_2 + R_3 + \cdots$$
The combined rate of multiple workers is the sum of their individual rates.
$R_{\text{total}}$=Combined work rate of all workers(jobs per unit time)
$R_1, R_2, \ldots$=Individual work rates(jobs per unit time)
Two workers with times $t_1$ and $t_2$
→Combined time = $\frac{t_1 \cdot t_2}{t_1 + t_2}$
A works for $a$ days, then B joins for $b$ days
→$\frac{a}{t_1} + \frac{b}{t_2} = 1$ (total work = 1 job)
Rates Add, Not Times: If A takes 6 days and B takes 12 days, the combined rate is $\frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$, so they finish in 4 days together.
LCM Shortcut: To avoid fractions, set total work = LCM of individual times. For A (6 days) and B (12 days), total work = 12 units. A's rate = 2 units/day, B's rate = 1 unit/day. Combined = 3 units/day → 4 days.
Sequential Work: If A works alone for some days and then B joins, split the work into two parts: A's solo contribution plus the combined contribution.
Efficiency compares how fast two workers are relative to each other. If A is twice as efficient as B, A's rate is double B's rate, which means A takes half the time B takes for the same job.
$$\frac{\text{Efficiency of A}}{\text{Efficiency of B}} = \frac{R_A}{R_B} = \frac{T_B}{T_A}$$
Efficiency is directly proportional to rate and inversely proportional to time.
$R_A, R_B$=Work rates of A and B(jobs per unit time)
$T_A, T_B$=Individual completion times(days, hours, etc.)
Efficiency Ratio = Inverse Time Ratio: If A takes 10 days and B takes 30 days, A is 3 times as efficient as B ($\frac{30}{10} = 3$).
Wages Split by Efficiency: If A and B work together, wages are divided in the ratio of their efficiencies (or their work contributions).
Man-Days Concept: Total work = number of workers × days. If 5 workers finish in 12 days, total work = 60 man-days. If 3 workers are assigned, they take 20 days.
Pipes and Cisterns
Pipe and cistern problems are identical to work problems, with one twist: a drain pipe has a negative rate. A fill pipe adds water (positive rate), while a drain pipe removes water (negative rate).
$$R_{\text{net}} = R_{\text{fill}} - R_{\text{drain}}$$
The net rate is the sum of all fill rates minus all drain rates.
$R_{\text{net}}$=Net filling rate of the system(tanks per unit time)
$R_{\text{fill}}$=Rate of fill pipe(s)(tanks per unit time)
$R_{\text{drain}}$=Rate of drain pipe(s)(tanks per unit time)
$R_{\text{net}} < 0$
→The tank will never fill — it drains faster than it fills.
$R_{\text{net}} = 0$
→Fill rate equals drain rate — the water level stays constant.
Fill Pipe as Worker: A pipe that fills a tank in $t$ minutes has rate $\frac{1}{t}$ tanks/minute — exactly like a worker.
Drain Pipe is Negative Rate: A drain that empties in $d$ minutes has rate $-\frac{1}{d}$ tanks/minute. Subtract it from fill rates.
Alternating Pipes: If pipes open/close in rotation, calculate work done per cycle (e.g., Pipe A open 3 min + Pipe B open 2 min = one cycle) and track the tank level.
Speed, Distance, and Time
The relationship between speed, distance, and time is the backbone of all motion problems. It's a simple three-variable equation that you rearrange depending on what you need to find.
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$
Speed equals distance divided by time. Rearrange to find distance ($d = s \times t$) or time ($t = \frac{d}{s}$).
$s$=Speed (rate of motion)(km/h, m/s, mph)
$d$=Distance travelled(km, m, miles)
$t$=Time taken(hours, seconds, minutes)
Three Forms: Memorize all three: $s = \frac{d}{t}$, $d = s \times t$, $t = \frac{d}{s}$. The triangle (S on top, D and T below) helps remember which to divide.
Constant Speed Assumption: These formulas assume constant speed throughout the journey. If speed changes, split the journey into segments.
Key Unit Conversions
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$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$
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$1 \text{ m/s} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$
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$1 \text{ km} = 1000 \text{ m}$
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$1 \text{ hour} = 60 \text{ min} = 3600 \text{ s}$
Average Speed
Average speed is NOT the average of two speeds. It is always total distance divided by total time — this is the single most important fact to remember.
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
Average speed weights each speed by the time spent at that speed — longer times have more influence.
$\text{Total Distance}$=Sum of all distances covered in each segment(km, m)
$\text{Total Time}$=Sum of all times for each segment(hours, seconds)
The Universal Rule: No matter how many segments, always compute $\frac{\sum d_i}{\sum t_i}$. Never just average the speeds.
Equal Distances Shortcut: When two equal distances are covered at speeds $a$ and $b$, the average speed is $\frac{2ab}{a+b}$ (harmonic mean, not arithmetic mean).
Equal Times Shortcut: When two equal time intervals are spent at speeds $a$ and $b$, the average speed IS $\frac{a+b}{2}$ (simple average — this is the rare case where averaging works).
Relative Speed and Trains
Relative speed is the effective speed between two moving objects. When objects move in opposite directions, their speeds add. When they move in the same direction, subtract the slower from the faster.
$$\begin{cases} s_{\text{rel}} = s_1 + s_2 & \text{(opposite directions)} \\ s_{\text{rel}} = |s_1 - s_2| & \text{(same direction)} \end{cases}$$
Relative speed depends on direction — objects approaching each other close the gap faster.
$s_{\text{rel}}$=Relative speed between two objects(km/h, m/s)
$s_1, s_2$=Individual speeds of the two objects(km/h, m/s)
Opposite Directions: Two cars at 50 km/h and 40 km/h heading toward each other close the gap at 90 km/h.
Same Direction: A faster car at 70 km/h chasing a slower car at 50 km/h closes the gap at only 20 km/h.
Meeting Time: Time to meet = initial gap $\div$ relative speed. Time to overtake = gap $\div$ relative speed (same direction).
Train problems add one extra layer: the train's own length counts as distance. When a train crosses a pole, it travels its own length. When it crosses another train, it travels the sum of both lengths.
$$t = \frac{L_1 + L_2}{s_{\text{rel}}}$$
Time to cross = total length to cover divided by relative speed.
$L_1$=Length of the first train(metres)
$L_2$=Length of the second train (or 0 for a pole/person)(metres)
$s_{\text{rel}}$=Relative speed (convert to m/s first)(m/s)
Crossing a pole or a person
→$L_2 = 0$, so time = $\frac{L_1}{s}$.
Crossing a platform
→Distance = train length + platform length.
Always Convert to m/s: Train lengths are in metres but speeds are usually given in km/h. Multiply by $\frac{5}{18}$ to convert.
Crossing vs Overtaking: Two trains moving in the same direction overtake each other; in opposite directions, they cross each other.
Platform Length: When a train crosses a platform, the total distance = train length + platform length, because the train's last carriage must clear the platform's far end.
Boats and Streams
A boat's effective speed depends on whether it moves downstream (with the current) or upstream (against the current). The stream adds to the boat's speed going downstream and subtracts going upstream.
$$s_{\text{down}} = b + c \qquad s_{\text{up}} = b - c$$
Downstream speed is boat speed plus stream speed; upstream is boat speed minus stream speed.
$s_{\text{down}}$=Speed of the boat going downstream(km/h)
$s_{\text{up}}$=Speed of the boat going upstream(km/h)
$b$=Speed of the boat in still water(km/h)
$c$=Speed of the stream/current(km/h)
Finding Boat and Stream Speeds: If you know $s_{\text{down}}$ and $s_{\text{up}}$: boat speed $b = \frac{s_{\text{down}} + s_{\text{up}}}{2}$ and stream speed $c = \frac{s_{\text{down}} - s_{\text{up}}}{2}$.
Time for Round Trip: Total time = $\frac{d}{b+c} + \frac{d}{b-c}$ where $d$ is the one-way distance.
Stream Speed Must Be Less Than Boat Speed: If $c \geq b$, the boat cannot move upstream ($s_{\text{up}} \leq 0$).
Circular Track Problems
On a circular track, two runners starting from the same point meet whenever the faster runner gains exactly one full lap on the slower one. The time to meet depends on the relative speed and the track length.
$$t_{\text{meet}} = \frac{L}{|s_1 - s_2|}$$
First meeting time equals track length divided by the relative speed (same direction).
$t_{\text{meet}}$=Time until they meet for the first time(seconds)
$L$=Length of the circular track(metres)
$s_1, s_2$=Speeds of the two runners(m/s)
Opposite directions
→Use $s_1 + s_2$ instead of $|s_1 - s_2|$: $t_{\text{meet}} = \frac{L}{s_1 + s_2}$.
$n$-th meeting
→Time = $n \times t_{\text{meet}}$ (each meeting takes the same time interval).
Same Direction: The faster runner must gain one full lap to meet. Relative speed = $|s_1 - s_2|$.
Opposite Direction: They approach each other, so relative speed = $s_1 + s_2$. They meet more quickly.
Starting at Different Points: If they start $k$ metres apart, first meeting time = $\frac{k}{s_{\text{rel}}}$ (same direction) or $\frac{L - k}{s_{\text{rel}}}$ (opposite direction, depending on which gap you measure).