Solving Rate, Distance, and Scheduling Problems

Good to see you again. In the previous lesson, you treated percentage, ratio, average, and mixture questions as representation problems: identify the correct total, convert ratios to scaled quantities, and preserve the relevant component. The same discipline now applies to motion and work questions.

Here, the key object is a rate: distance per unit time for motion, or fraction of a job per unit time for work. By the end of this lesson, you should be able to set up and solve standard GATE General Aptitude questions on work rates, speed–distance–time, and schedules in which workers or pipes begin, stop, or change over time. We will use the same approach that helps prevent negative-marking errors: establish units and the governing quantity before calculating.

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Rates: one structure behind travel and work

A rate answers the question: how much changes in one unit of time?

Situation	Quantity completed	Rate
A car travels	distance	km per hour or m per second
A worker completes a task	work	fraction of job per day
A pipe fills a tank	tank volume	fraction of tank per minute
A leak empties a tank	negative volume	negative fraction of tank per minute

The common relationship is:

$$\text{Quantity completed}=\text{Rate}\times \text{Time}$$

For travel:

$$D=S\times T$$

For work:

$$W=R\times T$$

where $R$ is the amount of work completed per unit time.

The Speed, Distance and Time Formula graphic is a useful memory aid, but do not let it replace unit checking.

For a work problem, if A completes a full job in $n$ days, set the whole job to $1$. A's daily work rate is then:

$$R_A=\frac{1}{n}$$

If A can finish in $12$ days, A does $\frac{1}{12}$ of the job per day. This is often called efficiency, but in calculations it is best viewed simply as a rate.

The most important habit for both topics is this:

Never add times or rates until their units and meanings match.

For example, $60$ km per hour cannot be directly combined with $400$ metres. Similarly, a pipe filling $\frac{1}{6}$ of a tank per hour and a leak emptying $\frac{1}{12}$ of a tank per hour must be treated as positive and negative rates of the same tank.

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Speed, distance, and time: set units before using the formula

The governing equations are:

$$S=\frac{D}{T}$$ $$D=S\times T$$ $$T=\frac{D}{S}$$

They are elementary, but GATE questions create difficulty by mixing units, changing speeds, or describing two moving objects.

Unit conversion: a high-value accuracy check

The conversions worth knowing are:

$$1\text{ km per hour}=\frac{5}{18}\text{ m per second}$$ $$1\text{ m per second}=\frac{18}{5}\text{ km per hour}$$

The direction matters:

• Convert km per hour to m per second by multiplying by $\frac{5}{18}$.
• Convert m per second to km per hour by multiplying by $\frac{18}{5}$.

Suppose a person crosses $600$ m in $5$ minutes.

Convert time:

$$5\text{ minutes}=300\text{ seconds}$$

Then:

$$S=\frac{600}{300}=2\text{ m per second}$$

If the answer is required in km per hour:

$$2\times \frac{18}{5}=7.2\text{ km per hour}$$

A quick plausibility check helps: walking speed around $7$ km per hour is possible; $72$ km per hour clearly signals a conversion error.

Average speed is not usually the average of speeds

The definition is always:

$$\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$$

Suppose a vehicle travels from A to B at $20$ km per hour and returns along the same route at $30$ km per hour. The distances are equal, but the times are not. Let each one-way distance be $d$.

$$\text{Total distance}=2d$$ $$\text{Total time}=\frac{d}{20}+\frac{d}{30}$$

Therefore:

$$\text{Average speed}=\frac{2d}{\frac{d}{20}+\frac{d}{30}}=24\text{ km per hour}$$

The arithmetic mean, $25$ km per hour, is wrong because the vehicle spends more time at the slower speed.

For two equal distances only, you may use:

$$\text{Average speed}=\frac{2xy}{x+y}$$

where $x$ and $y$ are the two speeds. This shortcut is a consequence of total distance divided by total time, not a replacement for it. Do not use it when distances differ or when the speeds apply for equal times rather than equal distances.

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Relative speed: track the gap, not each journey separately

When two objects move, the useful quantity is often the separation between them.

• When they move toward each other, the gap closes at the sum of their speeds.
• When a faster object catches a slower object moving in the same direction, the gap closes at the difference of their speeds.

Moving toward each other

Two buses start $600$ km apart and travel toward each other at $40$ km per hour and $60$ km per hour.

Their gap reduces at:

$$40+60=100\text{ km per hour}$$

Hence, the meeting time is:

$$\frac{600}{100}=6\text{ hours}$$

The logic is not merely “add opposite-direction speeds.” Over every hour, the first bus covers $40$ km and the second covers $60$ km, jointly removing $100$ km of separation.

Catch-up in the same direction

A runner has a $800$ m head start and moves at $10$ km per hour. A cyclist follows at $100$ km per hour.

The cyclist gains distance at:

$$100-10=90\text{ km per hour}$$

Convert this to metres per second:

$$90\times \frac{5}{18}=25\text{ m per second}$$

Therefore, catch-up time is:

$$\frac{800}{25}=32\text{ seconds}$$

There must be an initial gap for a same-direction catch-up question. If two objects leave the same point at the same time in the same direction, with unequal speeds, they separate immediately rather than meet later.

Early and late arrival problems: use the difference in travel times

A commuter travels the same route at $60$ km per hour and arrives $10$ minutes late. At $80$ km per hour, the commuter arrives $5$ minutes early. Find the distance.

The two arrival times differ by:

$$10+5=15\text{ minutes}$$

The slower-speed journey takes longer, so:

$$\frac{D}{60}-\frac{D}{80}=\frac{15}{60}$$ $$D\left(\frac{1}{60}-\frac{1}{80}\right)=\frac{1}{4}$$ $$D\left(\frac{1}{240}\right)=\frac{1}{4}$$ $$D=60\text{ km}$$

The most common interpretation mistake is to use $10-5=5$ minutes. “Ten minutes late” and “five minutes early” are on opposite sides of the scheduled arrival time, so the two actual arrival times are $15$ minutes apart.

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Work-rate problems: model a job as one complete unit

For work problems, choose one of two compatible representations:

1. Fraction method: treat the total job as $1$.
2. LCM method: choose a convenient total number of work units.

The fraction method is the most general. The LCM method can reduce arithmetic when completion times are integers.

Combined work: add productive rates

If A completes a job in $12$ days and B completes it in $18$ days:

$$R_A=\frac{1}{12}$$ $$R_B=\frac{1}{18}$$

Working together, their rate is:

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

So the completion time is:

$$\frac{1}{\frac{5}{36}}=\frac{36}{5}=7.2\text{ days}$$

The reciprocal appears because a rate of $\frac{5}{36}$ jobs per day means that one whole job takes $\frac{36}{5}$ days.

With the LCM method, choose total work as $36$ units:

• A's rate is $3$ units per day.
• B's rate is $2$ units per day.
• Together they complete $5$ units per day.

Thus:

$$\frac{36}{5}=7.2\text{ days}$$

Both methods are equivalent. Use fractions if the question includes partial work, joins, exits, or leaks. Use work units when the LCM makes all daily rates clean integers.

Negative work: subtract an opposing rate

A pipe fills a tank in $20$ minutes. A leak empties a full tank in $60$ minutes.

The pipe's rate is:

$$\frac{1}{20}$$

The leak's rate is:

$$-\frac{1}{60}$$

The net rate is:

$$\frac{1}{20}-\frac{1}{60}=\frac{2}{60}=\frac{1}{30}$$

So the tank fills in:

$$30\text{ minutes}$$

A useful sanity check: because the leak opposes filling, the answer must be more than $20$ minutes. Any answer below $20$ minutes has the wrong sign.

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Basic scheduling: divide the timeline whenever the active set changes

In this lesson, “scheduling” means rate problems in which the participants or conditions change over time: a worker joins late, a pipe is closed, a leak remains open for only part of the process, or shifts alternate. This is different from CPU scheduling, which you will study later in Operating Systems.

The central rule is:

Whenever the set of active workers, pipes, or machines changes, begin a new time interval.

Use this procedure:

1. Put time $0$ at the start of the process.
2. Mark every event: a worker starts, stops, joins, or leaves.
3. For each interval, write only the rates that are active in that interval.
4. Compute work completed as rate multiplied by interval length.
5. Add completed work and check whether the job finishes before the next event.

Consider this schedule:

• Pipe P fills a tank in $6$ hours.
• Pipe Q fills it in $9$ hours.
• Pipe R empties it in $12$ hours.
• P and R are open for the first $4$ hours.
• Then P closes and Q opens.
• R remains open for another $6$ hours, then closes.
• Q remains open until the tank is full.

Interval	Active pipes	Net rate	Duration	Work completed
First phase	P and R	$\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$	$4$ hours	$\frac{1}{3}$
Second phase	Q and R	$\frac{1}{9}-\frac{1}{12}=\frac{1}{36}$	$6$ hours	$\frac{1}{6}$
Final phase	Q only	$\frac{1}{9}$	unknown	remaining work

After $10$ hours, total work completed is:

$$\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$$

So half the tank remains. Q alone fills half the tank in:

$$\frac{\frac{1}{2}}{\frac{1}{9}}=\frac{9}{2}=4.5\text{ hours}$$

The total fill time is:

$$4+6+4.5=14.5\text{ hours}$$

Notice what prevents errors here: we did not average rates across the whole process. We calculated each interval using its actual active configuration.

Worker-count scheduling and product constancy

If all workers have equal efficiency and work the same number of hours per day, then for the same total job:

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

More generally:

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

Suppose $25$ equally efficient workers can complete a job in $8$ days. How many are needed to complete the same job in $20$ days, with the same daily hours?

$$25\times 8=M\times 20$$ $$M=10$$

The number of workers falls because the available time increases. This inverse relationship is the conceptual check.

Be careful about what the question asks. If $52$ workers were expected to finish in $10$ days, but $12$ are reassigned, then $40$ workers remain. The new completion time is:

$$52\times 10=40\times D$$ $$D=13\text{ days}$$

If asked for the extra days, the answer is not $13$. It is:

$$13-10=3\text{ days}$$

This is a classic GATE interpretation trap: the equation is correct, but the requested quantity is different.

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A compact decision and checking routine

These questions can be good early attempts in a GATE paper when the setup is visible. Before committing, apply a short check.

For speed and distance

• Are distance and speed in compatible units?
• Is the question asking for a total distance, a gap, or a train length?
• For average speed, did you use total distance divided by total time?
• For two moving objects, is the separation closing or increasing?
• For “early” and “late,” did you add the time differences across the schedule?

For work and scheduling

• Did you set the full job or full tank to one consistent unit?
• Does each worker or pipe rate have the correct sign?
• Did you split the timeline at every start, stop, or switch event?
• Is the final answer plausible? A leak cannot make a tank fill faster; fewer identical workers cannot finish the same job in less time.
• Does the question ask for total time, remaining time, extra time, or work completed?

For your error log, label a missed question precisely:

• Unit error: km per hour and metres, or days and hours, were mixed.
• Rate-direction error: added a leak or used the wrong relative speed.
• Timeline error: applied one rate across intervals with different active workers.
• Interpretation error: calculated total days when the question asked for extra days.
• Calculation error: correct model, incorrect fraction arithmetic.

That classification will make your later reattempts much more effective than merely recording “time and work weak.”

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Key takeaways

Speed–distance–time and work-rate questions share one structure: quantity completed equals rate multiplied by time.

• Use consistent units before applying $D=S\times T$.
• Compute average speed as total distance divided by total time.
• Use the sum of speeds for objects approaching each other and the difference for a genuine same-direction catch-up.
• Model a full job as $1$, so a worker completing it in $n$ days has rate $\frac{1}{n}$.
• Add helpful rates and subtract opposing rates such as leaks or emptying pipes.
• In scheduling problems, divide the timeline whenever the active set changes.
• Use worker-days product constancy only when the job, working hours, and worker efficiencies are appropriately comparable.

In the next lesson, you will shift from constructing rates to extracting exactly the required quantities from tables, charts, and short data sets—another high-return area where careful reading is often more important than difficult arithmetic.