Time and Work Made Easy: The LCM (Total Work) Method

Most students solve time-and-work questions with fractions like 1/12 + 1/15. That works, but it is slow and easy to get wrong under exam pressure. The LCM method turns every question into simple whole-number arithmetic.

The idea: assume the total work

1. Take the LCM of the days given. Call that the total work (in “units”).
2. Each person’s units per day = total work ÷ their days.
3. Add or subtract the daily units as the question needs, then divide.

Example 1: two people together

A finishes a job in 10 days and B in 15 days. How long together?

• LCM(10, 15) = 30 units of work.
• A does 30 ÷ 10 = 3 units/day; B does 30 ÷ 15 = 2 units/day.
• Together: 5 units/day → 30 ÷ 5 = 6 days.

Example 2: finding one person’s time

A and B together take 8 days; A alone takes 12 days. How long does B take alone?

• LCM(8, 12) = 24 units.
• A + B = 3 units/day; A = 2 units/day, so B = 1 unit/day.
• B alone: 24 ÷ 1 = 24 days.

Example 3: someone leaves part-way

A takes 15 days and B 20 days. They work together for 6 days, then A leaves. How long does B take to finish?

• LCM = 60 units; A = 4/day, B = 3/day, together 7/day.
• In 6 days: 42 units done, 18 left.
• B alone: 18 ÷ 3 = 6 more days.

Efficiency questions

“A is twice as efficient as B” means A does 2 units for every 1 that B does. You don’t need the LCM here: use the efficiencies as the daily units.

Example: together they take 14 days. Daily work = 2 + 1 = 3 units, so total = 3 × 14 = 42 units. A alone: 42 ÷ 2 = 21 days.

“B is 50% more efficient than A” means B does 1.5 times as much per day, so B takes less time: 18 ÷ 1.5 = 12 days if A takes 18.

Men-days (work equivalence)

For “M men in D days” questions, total work = men × days (× hours per day), and it stays constant.

Example: 10 men working 8 hours a day finish in 12 days. Total = 10 × 8 × 12 = 960 man-hours. For 8 men working 10 hours a day: 960 ÷ 80 = 12 days.

Pipes and cisterns: same method, with signs

A filling pipe adds units; an emptying pipe or leak subtracts them.

Example: A fills a tank in 12 hours, B in 15 hours, and C empties it in 20 hours. All open together:

• LCM(12, 15, 20) = 60 units (the tank).
• A = +5, B = +4, C = −3 → net +6 units/hour.
• 60 ÷ 6 = 10 hours.

Leak questions: “A pipe fills in 8 hours but takes 10 hours because of a leak.” LCM = 40: the pipe fills 5 units/hour, the actual rate is 4, so the leak removes 1 unit/hour and empties the full tank in 40 hours.

Alternate days: work in pairs

A takes 8 days and B 12 days, working on alternate days starting with A.

• LCM = 24 units; A = 3, B = 2; one pair of days = 5 units.
• After 4 pairs (8 days): 20 units done, 4 left.
• Day 9 is A’s: 3 units, leaving 1. B needs 1 ÷ 2 = half a day.
• Total: 9.5 days.

Always check who works on the final day: the answer changes if B starts first.

Wages

Wages are shared in the ratio of work done, which is the ratio of daily units if everyone works the same number of days. If A does 2 units/day and B 1 unit/day, A gets 2/3 of the payment.

Quick checklist

Question says	Do this
Days for each person	LCM → units per day
“Twice as efficient”	Use 2 : 1 as units per day
Men, days, hours	Total = men × days × hours, constant
Pipe empties / leak	Subtract its units
Alternate days	Count work per pair of days, check the last day

Practise with our time and work, pipes and cisterns set. Every answer there was calculated by a script to make sure it’s exact.