Maths olympiad question · P5–P6★★★★ Expert
Work problems: olympiad question 11
An expert question on work problems for P5–P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P3 to P6 papers; harder ones have someone leaving partway, people taking turns, or cows grazing a field that keeps growing.
The question
An estate runs two repair jobs at once.
Job 1 belongs to crew A, 12 workers, who would take 18 days on their own; job 2 to crew B, 4 workers, who would take 31 days on their own.
Both start the same day, but job 2's site is closed for the first 3 days (crew A works as usual).
Later crew A sends 3 workers over to help crew B, and the two jobs finish on the same day.
Every worker does the same amount each day, with no breaks.
For how many days did crew A work before the move?
Show the hints (3)
- Hint 1: Call one worker's day 1 part. How many parts are the two jobs together?
- Hint 2: Apart from crew B's 3 shut days, every worker works from start to finish. First find how many days the jobs took.
- Hint 3: The jobs are 216 + 124 = 340 parts; crew B's closure lost 4 × 3 = 12, so the jobs took 22 days. Job 1: crew A has 12 workers, then 9.
Worked solution
Have a go first, then read the solution step by step.
- 1.Let one worker's day be 1 part. Job 1 = 12 × 18 = 216 parts, job 2 = 4 × 31 = 124.
- 2.Moving workers changes where, not how many. From start to finish all 16 work every day, except crew B's 4 for the 3 shut days: 4 × 3 = 12 parts fewer.
- 3.So (216 + 124 + 12) ÷ 16 = 352 ÷ 16 = 22 days: both jobs finish on day 22.
- 4.Job 1: had crew A been 9 for all 22 days, they would do 9 × 22 = 198 parts, 18 short of 216. The shortfall is what the extra 3 did before the move.
- 5.So the move came after 18 ÷ 3 = 6 days.
- 6.Check: job 1, 12 × 6 + 9 × 16 = 216 = 216 ✓; job 2, 4 × 19 + 3 × 16 = 124 = 124 ✓
- ⚠️Common mistake: forgetting crew B's 3 shut days and finding the finish day as (216 + 124) ÷ 16; or counting the 3 moved workers' days on job 1.
Answer: 6 days
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