Work questions are the quantitative section's most procedural family. "A can do a job in 12 days and B in 18" is not a problem to be intuited; it is a template to be applied, and the template never changes. Learn four setups — worker-days, combined rates, pipes, and partway joins — and you can convert a whole block of questions into mechanical arithmetic.
The classic failure here is adding times. Two workers taking 12 and 18 days do not take 30 days together, and they do not take 15. The reason is worth understanding once, because it is the same reason average speed is not the average of speeds.
Convert every worker into a rate: the fraction of the job they finish in one unit of time.
When workers act together, their rates add, because in one day they collectively finish the sum of their daily shares.
For two workers the shortcut T = ab ÷ (a + b) is worth memorising: 12 and 18 days give (12 × 18) ÷ 30 = 7.2 days, and the answer is always less than the smaller of the two times — a one-second sanity check that kills most wrong options.
If the work is measured in people rather than days, the total work is constant:
Fifteen workers finish a job in 20 days, so the job is 300 worker-days; 25 workers need 300 ÷ 25 = 12 days. If the question mentions hours per day, convert to worker-hours first: 15 workers × 20 days × 8 hours = 2,400 worker-hours.
Add the fractions, then invert. Three workers follow the same rule: 1 ÷ T = 1 ÷ a + 1 ÷ b + 1 ÷ c, so 6, 12 and 12 days give 1/6 + 1/12 + 1/12 = 1/3, which is 3 days.
An inlet adds its rate; an outlet subtracts. A pipe filling in 6 hours and a drain emptying in 9 hours have a net rate of 1/6 − 1/9 = 1/18, so the tank fills in 18 hours — slower than the inlet alone, which is the point of the question.
1/12 + 1/18 = 3/36 + 2/36 = 5/36, so T = 36/5 = 7.2 days. The shortcut agrees: 216 ÷ 30 = 7.2. Sanity check: less than 12, as it must be.
A's four days produce 4/10 = 0.4 of the job, leaving 0.6. Working together the rate is 1/10 + 1/15 = 1/6, so the remaining work takes 0.6 × 6 = 3.6 days. Total: 4 + 3.6 = 7.6 days.
Total work = 12 × 15 = 180 worker-days, so nine workers need 180 ÷ 9 = 20 days. Fewer workers, more days — inverse proportion.
Let B's rate be r, so A's is 2r and the total is 3r = 1/8, giving r = 1/24. B alone would need 24 days and A alone 12 days. Check: 1/12 + 1/24 = 3/24 = 1/8.
A can paint a room in 6 days and B in 12. Together?
Fifteen workers complete a job in 20 days. How many days for 25 workers at the same rate?
An inlet fills a tank in 4 hours and a drain empties it in 6 hours. With both open, how long to fill it?
A does a job in 10 days and B in 15. They work together for 3 days. What fraction of the job remains?
Without looking back: how do you turn "a days alone" into a rate, and what happens to rates when workers act together?
The answer: the rate is 1 ÷ a jobs per day, and rates add when workers act together.
A worker who needs a days alone has a rate of 1 divided by jobs per day. When workers act together their rates . For two workers taking 12 and 18 days, the combined time is days, and the total work in a worker-days problem is workers days.
A question reads: "Pipe A fills a tank in 8 hours. Pipe B fills it in 12 hours, and a drain C empties it in 24 hours. All three are opened. How long until the tank is full?" Solve it and give the sanity check.
Net rate = 1/8 + 1/12 − 1/24. Over a common denominator of 24: 3/24 + 2/24 − 1/24 = 4/24 = 1/6. So the tank fills in 6 hours.
The sanity check is a bound test: without the drain, pipes A and B together fill the tank in 4.8 hours (1/8 + 1/12 = 5/24), so the answer must be greater than 4.8 and less than 8 (the best single pipe). Six hours satisfies both bounds, which eliminates any option outside that window without a single extra calculation.
Rates are done. The next lessons turn to counting and probability — how many ways something can happen, and how likely a given outcome is — which is the last block of quantitative material with its own vocabulary.