Individual rate
Convert time into one-day work.
Rate = 1 / days
If A finishes in 10 days, A's rate is 1/10 work per day.
Quantitative Aptitude
Learn work rates, efficiency, combined work, alternate days, pipes, and man-days with fast LCM-based solving.
Time and work problems are solved by converting "time taken" into "work per unit time" (efficiency). Choose a convenient total work — usually the LCM of the given days — then express each worker or pipe as an integer rate per day. The same engine handles combined work, pipes and cisterns, men-and-days proportionality, and wage distribution (wages split in the ratio of efficiencies).
If a person completes work in N days, their one-day work is 1/N. Combined work is the sum of individual rates.
Convert time into one-day work.
If A finishes in 10 days, A's rate is 1/10 work per day.
Add work rates when people work together.
A in 10 days and B in 15 days together finish in 6 days.
LCM method avoids fractions. Take total work as LCM of days, then compute each person's daily efficiency.
Use the LCM of completion days as total work units.
For 10 and 15 days, total work = 30 units. A = 3 units/day, B = 2 units/day.
For same work, workers and days are inversely proportional.
6 men in 20 days equals 15 men in 8 days.
Memorize the high-frequency conversions and rate patterns so calculations become instant during timed tests.
| Completes in 5 days | 1/5 work/day |
| Completes in 10 days | 1/10 work/day |
| Completes in 20 days | 1/20 work/day |
| Completes in 25 days | 1/25 work/day |
Convert completion time into daily rate.
Add rates, then invert the result.
Use for workforce and shift questions.
If A is faster, A takes fewer days and has higher efficiency.
The recurring question patterns in this topic — know the shape of each pattern before you solve.
Two or more workers complete a job; find combined time via rate addition.
A: 10 days, B: 15 days → together 6 days.
Change crew size or days; work stays constant.
6 men in 20 days → 15 men in 8 days.
Some work is done for a few days; find what fraction is left.
A (12 d) and B (18 d) work 4 days → 5/9 done, 4/9 remains.
Inlet adds to the tank, outlet drains it; net rate decides fill time.
Fill 6 h, leak 12 h → net 1/12 per hour → 12 hours.
Money is split among workers in the ratio of their efficiencies.
A: 10 d, B: 15 d → efficiency ratio 3:2.
High-signal questions with full solution flow across difficulty patterns.
Classic exam problems worth internalizing — each one ships a complete step-by-step solution.
Convert fractional rates into clean daily units.
Calculate completed work in repeated cycles, then solve the remainder.
Use for pipes, leaks, and work undone.
Where most students lose marks — review these before you sit the test.
You have mastered the theory. Now put it under pressure with a focused topic test and instant explanations.
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