Time and Work Formulas, Shortcuts, and Tricks

Time and Work problems reduce to one fraction: a worker who finishes a job in n days completes 1/n of it each day. Master that fraction, and every variant (combined rates, wages splits, pipes and cisterns) follows from the same arithmetic.

The Core Formula: One Day’s Work

Rate and time are inverses. A worker who takes 10 days completes 1/10 of the job per day; one who takes 15 days completes 1/15 per day. When workers combine, add their rates. The sum of rates, inverted, gives the combined time.

Three base rules follow directly:

• If A finishes in n days, A’s one-day work = 1/n.
• Two workers together: add individual rates, then invert the sum. If A takes 10 days and B takes 15 days, combined rate = 1/10 + 1/15 = 3/30 + 2/30 = 5/30, combined time = 30/5 = 6 days.
• If A is n times as efficient as B, A’s rate is n times B’s rate, and A’s time is B’s time divided by n.

The fraction method above is correct. For problems with three or more workers or unequal denominators, the LCM shortcut below avoids fraction arithmetic entirely.

The LCM Shortcut

The LCM shortcut is the standard approach for multi-worker aptitude problems. Instead of working with unequal fractions, it sets the total work as the LCM of all individual times and converts every rate to a whole number. Arithmetic errors in campus tests almost always come from fraction addition; the LCM method removes that risk.

Steps:

• Step 1: Set total work = LCM of all individual completion times.
• Step 2: Each worker’s daily output = total work divided by that worker’s time.
• Step 3: Combined daily output = sum of all individual outputs. Time to finish = total work divided by combined output.

Worked example (A: 10 days, B: 12 days):

• LCM(10, 12) = 60 units of total work.
• A’s output: 60 / 10 = 6 units per day. B’s output: 60 / 12 = 5 units per day.
• Combined output: 6 + 5 = 11 units per day.
• Time to finish 60 units: 60 / 11 days, approximately 5.45 days.

Worked example (A: 9 days, B: 18 days):

• LCM(9, 18) = 18 units of total work.
• A: 18 / 9 = 2 units per day. B: 18 / 18 = 1 unit per day. Combined: 3 units per day.
• Time: 18 / 3 = 6 days.

IndiaBix’s Time and Work practice set includes 60-plus problems across these question types, including three-worker variants where the LCM method’s advantage over fraction tracking is most visible.

For two workers with simple denominators, both methods take similar time. For three workers or messy denominators, set total work to LCM once and work entirely with whole numbers.

Efficiency Ratios and Wages

Efficiency Ratios

When a problem states “A is k times as efficient as B,” that defines the rate ratio directly. A’s rate equals k times B’s rate. Because rate and time are inverses, A’s time equals B’s time divided by k.

Worked example (A is 3 times as efficient as B; B alone takes 24 days):

• A’s time alone: 24 / 3 = 8 days.
• Set up LCM: LCM(8, 24) = 24 units total work.
• A’s output: 24 / 8 = 3 units per day. B’s output: 24 / 24 = 1 unit per day.
• Combined output: 3 + 1 = 4 units per day.
• Time together: 24 / 4 = 6 days.

Quick-check: A does 3/4 of the work, B does 1/4. A is 3 times as efficient. Both checks hold.

Wages

Wages in Time and Work problems split in proportion to work done, not time worked. The LCM method gives each worker’s unit total; that ratio becomes the wage ratio.

Steps:

• Step 1: Find total work via LCM method.

• Step 2: Multiply each worker’s daily rate by the days they actually worked to get their unit total.

• Step 3: Divide the total wage in the ratio of those unit totals.

• Worked example: A finishes in 10 days, B in 15 days; A works 1 day then B finishes; total wage ₹1,000.

• LCM(10, 15) = 30 units total work.

• A’s rate: 30 / 10 = 3 units per day. B’s rate: 30 / 15 = 2 units per day.

• A works 1 day: contributes 3 units. Remaining: 30 - 3 = 27 units. B completes all 27.

• Wage ratio A : B = 3 : 27 = 1 : 9. Total parts = 10.

• B’s share: (9 / 10) × 1,000 = ₹900. A’s share: (1 / 10) × 1,000 = ₹100.

• Check: 900 + 100 = 1,000. ✓

A quick ratio check after computing takes two seconds and catches transposition errors before submission.

Man-Days: The Workforce-Change Method

When the number of workers changes, the man-days principle applies: total work in man-days stays constant. If M workers complete a job in D days, that job requires M × D man-days. Increase the workforce and the required days decrease proportionally; decrease it and days increase.

Formula: M1 × D1 = M2 × D2, assuming each worker works the same hours per day.

Worked example (12 men complete 1 task in 14 days; find the time for 28 men to complete 10 tasks):

• Total work for 1 task: 12 × 14 = 168 man-days.
• 28 men working on 1 task: 168 / 28 = 6 days.
• 10 tasks in sequence: 6 × 10 = 60 days.

When the problem also changes working hours, the full formula becomes M1 × D1 × H1 = M2 × D2 × H2, where H is hours per day. Add the hours dimension only when the problem explicitly states a change in daily hours.

Partial completion is another common variant: if work is half done, reduce the remaining man-days by half before applying the formula to find the new timeline.

Pipes and Cisterns

Pipes and cisterns problems are structurally identical to worker problems. A filling pipe adds to the net rate; an emptying pipe subtracts from it.

Net rate = (sum of all filling-pipe rates) - (sum of all emptying-pipe rates).

Worked example (P fills in 10 minutes, Q fills in 15 minutes, R empties in 12 minutes; all three open together):

• LCM(10, 15, 12) = 60 litres total capacity (convenient working unit).
• P rate: 60 / 10 = 6 litres per minute (filling). Q rate: 60 / 15 = 4 litres per minute (filling). R rate: 60 / 12 = 5 litres per minute (emptying).
• Net rate: 6 + 4 - 5 = 5 litres per minute.
• Time to fill 60 litres: 60 / 5 = 12 minutes.

With only P and Q open and R closed, the fill time is 60 / (6 + 4) = 6 minutes. Opening R triples the fill time in this example.

When only the emptying pipe R is open and the tank is full, it drains in 60 / 5 = 12 minutes. Any question that adds filling pipes while R runs just adds those rates before dividing.

The sign rule is everything: treat each pipe as a worker, mark its sign (positive for filling, negative for emptying), sum the signed rates, then invert.

These five question types (combined rate, finding an unknown rate, departure problems, wages split, and man-days) cover the majority of Time and Work questions in campus placement aptitude tests. For step-by-step worked examples across all five, FACE Prep’s Time and Work solved questions walks through each pattern. For context on how the full aptitude test is structured and what cutoff scores look like, see the campus placement evaluation test guide.

The rate-addition logic that runs through every section here (express each component as a unit fraction, add the signed rates, invert the sum) is the same decomposition habit that well-structured AI tasks reward. TinkerLLM is a low-cost environment to apply that structured thinking to LLM problems; at ₹299, it’s a reasonable next step once aptitude drills are solid.