Combined work problems

Two painters, two taps, two typists — whenever more than one worker shares a single job, the everyday instinct is to average their times. That instinct is wrong, and knowing exactly why is one of the sharpest proportional-reasoning skills a Class 8 student can own. The fix is small but powerful: stop thinking in days and start thinking in the fraction of the job done per day — the work rate. Rates add when people work together; days never do. This idea sits directly inside the CBSE "Proportional Reasoning-2" chapter, where workers-and-days problems are a standard inverse-proportion context: more workers means less time, in exact inverse proportion. Master the rate trick here and the same reasoning carries into taps filling and draining a tank, pipes, and any "how long together" question you will meet through Class 10 and beyond.

The core insight

Convert "days to finish alone" into "fraction of the job done per day," add the individual rates to get the combined rate, then take the reciprocal of that sum to find the time taken together.

The core idea

Work problems feel like speed problems in disguise, and the disguise is deliberate — the same inverse-proportion habit that governs "more workers, fewer days" governs "faster combined rate, less combined time."

Rate, not days, is the additive quantity. If a person finishes a job alone in $n$ days, they complete a constant fraction of the job every day:

\text{rate} = \dfrac{1}{n} \text{ of the job per day}

where: $n$ is the number of days one person alone needs, and the rate is a fraction of one whole job completed in one day — never a number of days itself.

Rates add when people work together. If Person A alone takes $a$ days and Person B alone takes $b$ days, working together their daily output simply sums:

\text{combined rate} = \dfrac{1}{a} + \dfrac{1}{b}

where: $a$ and $b$ are the individual times alone (in days), and the combined rate is the fraction of the job the pair finishes together in one day.

Time is the reciprocal of the combined rate. If the combined rate equals $\dfrac{1}{T}$ , then the whole job — one complete unit of work — takes

T = \dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}} = \dfrac{ab}{a+b} \text{ days}

where: $T$ is the number of days needed when both work together, and $\dfrac{ab}{a+b}$ is the same relation written as a single fraction after combining $\dfrac{1}{a}+\dfrac{1}{b}$ . The pattern extends to any number of workers or machines: with individual rates $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dots$ , the combined daily rate is their sum, and the time taken together is the reciprocal of that sum. This is exactly the inverse-proportion logic of the syllabus — the amount of work is a fixed constant, so increasing the rate (more hands on deck) proportionally shrinks the time.

Figure

Two bars showing individual daily work fractions of one twelfth and one eighth stacking into a combined bar of five twenty fourths per day

Derivation of the combined-time formula

Derivation ACombined time from two individual times

The goal is to show, purely from the definition of "rate as fraction of the job per day" and the algebra of adding two fractions, that two people who alone take $a$ days and $b$ days will together finish the job in exactly $T = \dfrac{ab}{a+b}$ days, and that this single compact formula is nothing more than a reciprocal of a sum in disguise, so it never needs to be memorised separately from the two-step add-then-flip method.

Step 1. Write each person's daily output as a fraction of the whole job. Person A alone needs $a$ days, so in one day A completes $\dfrac{1}{a}$ of the job; Person B alone needs $b$ days, so B completes $\dfrac{1}{b}$ of the job in one day. These are the two work rates, each a pure fraction of "one job," with no reference to a physical unit like bricks or pages — that is exactly why they can be added while raw days cannot.

Step 2. Add the two rates to get the combined daily output. Working together on the same job, in one day A contributes $\dfrac{1}{a}$ and B contributes $\dfrac{1}{b}$ of the total, so together they complete $\dfrac{1}{a} + \dfrac{1}{b}$ of the job in that day. Using the common denominator $ab$ , this sum is $\dfrac{b}{ab} + \dfrac{a}{ab} = \dfrac{a+b}{ab}$ .

Step 3. Take the reciprocal to convert the daily fraction back into a time. If the pair completes $\dfrac{a+b}{ab}$ of the job each day, then completing the whole job (one full unit of work) takes $T$ days where $T \times \dfrac{a+b}{ab} = 1$ , so $T = \dfrac{ab}{a+b}$ days — exactly the reciprocal of the combined rate found in Step 2.

Step 4. Confirm the formula matches the "add, then flip" method directly: $\dfrac{1}{T} = \dfrac{1}{a}+\dfrac{1}{b}$ rearranges to $T = \dfrac{1}{\frac{1}{a}+\frac{1}{b}} = \dfrac{ab}{a+b}$ , so the single-formula shortcut and the step-by-step rate method are the same algebra, and either is acceptable in an exam as long as the reasoning (rates add, then reciprocate) is shown.

Worked example

Ravi can paint a fence alone in $12$ days. Meena can paint the same fence alone in $8$ days. If they work together, how many days will they take?

Step 1. Find each rate. Ravi: $\dfrac{1}{12}$ of the fence per day. Meena: $\dfrac{1}{8}$ of the fence per day.

Step 2. Add the rates using the LCM of $12$ and $8$ , which is $24$ : $\dfrac{2}{24} + \dfrac{3}{24} = \dfrac{5}{24}$ of the fence per day, together.

Step 3. Take the reciprocal: $T = \dfrac{1}{\frac{5}{24}} = \dfrac{24}{5} = 4.8$ days. Together, Ravi and Meena finish the fence in 4.8 days (about $4$ days and $19.2$ hours). Check: in $4.8$ days Ravi does $\dfrac{4.8}{12}=0.4$ of the job and Meena does $\dfrac{4.8}{8}=0.6$ ; $0.4+0.6=1$ , the whole job. ✓

Competitive stretch

$A$ and $B$ together complete a task in $6$ days; $A$ alone takes $10$ days. How many days would $B$ take alone?

Step 1. Combined rate: $\dfrac{1}{6}$ of the job per day. Step 2. $A$ 's rate alone: $\dfrac{1}{10}$ per day. Step 3. Subtract to isolate $B$ 's rate: $\dfrac{1}{6}-\dfrac{1}{10} = \dfrac{5}{30}-\dfrac{3}{30} = \dfrac{2}{30} = \dfrac{1}{15}$ . Step 4. $B$ 's rate is $\dfrac{1}{15}$ of the job per day, so $B$ alone takes 15 days. Check: $\dfrac{1}{10}+\dfrac{1}{15} = \dfrac{3}{30}+\dfrac{2}{30} = \dfrac{5}{30} = \dfrac{1}{6}$ ✓.

Common traps

Averaging the days instead of adding the rates. $12$ days and $8$ days do not average to the combined time; add $\dfrac{1}{12}+\dfrac{1}{8}$ , never $\dfrac{12+8}{2}$ .
Forgetting the final reciprocal. The sum of rates gives work done per day; the time taken is the reciprocal of that sum, not the sum itself.
Skipping a common denominator. Adding or subtracting the fractions without first finding the LCM of the individual times leads straight to a wrong combined fraction.
Adding a rate that should be subtracted. When one agent works against the others (for example a leak draining a tank the others are filling), that rate must be subtracted, not added — the "work backward" example above uses the same subtraction logic to find a missing individual time.

Ask the tutor: Want to try a three-worker combined-work problem, or one where a tap fills a tank while a leak drains it?

Why can we not just average 12 days and 8 days to get the combined time

Averaging treats days as if they add up directly, but a "day" is not a fixed chunk of work — it means different amounts for different people. Ravi's day is worth one twelfth of the fence and Meena's day is worth one eighth, a bigger chunk. Only the fractions of work per day are directly comparable and addable, because each is measured against the same whole job. Once you add those fractions and flip the result, you automatically get a time between the two individual times but weighted correctly, which is why together-time is always less than the faster worker's alone-time, not a simple midpoint.

Does the combined-time formula T equals ab over a plus b work for three or more workers

Not directly as written, because that compact form comes from adding exactly two fractions. The underlying method scales perfectly though: add every individual rate, one over a, one over b, one over c and so on, to get the combined daily rate, then take the reciprocal of that sum for the total time. If you insist on a single formula for three workers it becomes messier, so it is safer and clearer in an exam to always go through the rate-then-reciprocal steps rather than memorise a bigger formula.

What if one person stops working partway through the job

Then you track work done in stages using the same fraction-per-day idea. Suppose both work together for some days, contributing their combined rate each day, and then one person continues alone. Add up the fraction of the job completed in each stage and set the total equal to $1$ (the whole job) to solve for the unknown time. This is a natural extension of the same "fractions of work add up to one whole job" principle used in the worked examples, just applied over more than one stage.