Time and Work, Pipes and Cisterns

The key idea: convert "does the job in n days" into a rate of 1/n of the job per day. Rates of people working at the same time add; the time taken is 1 / (total rate).

Working together

• A finishes a job in 10 days and B in 15 days. Together: 1/10 + 1/15 = 1/6 of the job per day, so 6 days.
• A, B and C take 12, 15 and 20 days: 1/12 + 1/15 + 1/20 = 12/60 = 1/5, so 5 days.
• A and B together take 8 days and A alone takes 12. Then B's rate is 1/8 − 1/12 = 1/24: 24 days alone.

Working in shifts

A alone takes 20 days, B alone 30. A works alone for 5 days, then B joins. After 5 days, 5/20 = 1/4 of the job is done; the remaining 3/4 at the joint rate 1/20 + 1/30 = 1/12 takes 9 days. The job takes 14 days in all.

Man-days

If the work is fixed, (number of workers) × (days) × (hours per day) is constant:

M₁D₁H₁ = M₂D₂H₂ (and if the amount of work changes, divide each side by its work W).

10 workers take 12 days, so 15 workers take 10 × 12 / 15 = 8 days.

Pipes and cisterns

A filling pipe has a positive rate and a leak or emptying pipe a negative one. A pipe fills a tank in 6 hours and a leak empties it in 9 hours: 1/6 − 1/9 = 1/18, so with both open the tank fills in 18 hours.

from fractions import Fraction as F

def together(*days):
    """Days needed when workers with the given single-handed times work together (negative = emptying)."""
    return 1 / sum(F(1, d) for d in days)

print("A 10, B 15:", together(10, 15), "days")
print("A 12, B 15, C 20:", together(12, 15, 20), "days")
print("B alone:", 1 / (F(1, 8) - F(1, 12)), "days")
print("fill 6 h, leak 9 h:", together(6, -9), "hours")
remaining = 1 - F(5, 20)
print("shift problem:", 5 + remaining / (F(1, 20) + F(1, 30)), "days in total")

A 10, B 15: 6 days
A 12, B 15, C 20: 5 days
B alone: 24 days
fill 6 h, leak 9 h: 18 hours
shift problem: 14 days in total

Practice questions